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🚀The Ultimate Guide to pH, pOH, [H⁺] & [OH⁻] Numericals for FSC, Matric & MDCAT/ECAT | Master Chemistry Like a Pro with InamJazbi! 🔥📚
🔥📘Formulas on pH and pOH
⚡pH = – log [H⁺]
⚡pOH = – log [OH⁻]
⚡pH + pOH = 14
⚡ log [H⁺] = – pH of solution
⚡[H⁺] = 10⁻ᴾᴴ
⚡Kw = [H⁺] [OH⁻] = 1 x 10⁻¹⁴ M⁻²
🔥📘Numericals on pH and pOH
Q1. A solution of HCl has pH of 2.3. Calculate its pOH and [H⁺]? (Book)
Solution
Calculation of pOH
pH + pOH = 14
pH = 14 – pOH
pOH = 14 – 2.3 = 11.7
Calculation of [H⁺]
pH = – log [H⁺]
10^x = [H⁺]
10⁻ᴾᴴ = [H⁺]
10⁻²˙³ = [H⁺]
[H⁺] = 5.01 × 10⁻³
Alternate Method
pH of solution = – log [H⁺]
log [H⁺] = – pH of solution = − 2.3 (On rearranging the equation)
Since exponent should be the whole number, we add and subtract the whole number that is closest to and larger than the negative log.
log [H⁺] = − 2.3 = (−2.3 + 3) − 3 = (0.7 − 3)
Taking antilog on both sides
[H⁺] = antilog (0.7 − 3)
[H⁺] = antilog 0.7 × antilog − 3
[H⁺] = 5.01 × 10⁻³
[H⁺] = 5.01 × 10⁻³
Q2. Find pH, pOH, [OH⁻] and [H⁺] of 2.46 × 10⁻⁹ M KOH solution. (Book)
Solution
Ionization Equation
KOH ⇌ K⁺ + OH⁻
1 mol 1 mol 1 mol
2.46 ×10⁻⁹M 2.46 × 10⁻⁹ M 2.46 × 10⁻⁹ M
Calculation of [OH⁻]
Since KOH is strong base, it is completely ionized. Hence concentration of KOH is equal to concentration of OH⁻ ions
[OH⁻] = [KOH] = 2.46 × 10⁻⁹ M
Calculation of [H⁺]
[H⁺][OH⁻] = Kw
[H⁺] × 2.46 × 10⁻⁹ = 1 × 10⁻¹⁴
H⁺] = 1 × 10⁻¹⁴/2.46 × 10⁻⁹
[H⁺] = 4.07 × 10⁻⁶
Calculation of pH
pH = – log [H⁺]
pH = – log [4.07 × 10⁻⁶]
pH = – [log 4.07 + log 10⁻⁶] (∴ log ab = log a + log b)
pH = – [0.609 + (–6)] (∴ log 4.07 = 0.609 and log 10⁻⁶ = –6 )
pH = – [0.609 – 6)]
pH = 5.39
Calculation of pOH
pH + pOH = 14
pOH = 14 – pH
pOH = 14 – 5.39
pOH = 8.61
Q3. Calculate the pH and pOH of 0.001 M solution of nitric acid (HNO₃).
Solution
Ionization; HNO₃(aq) ⇌ H⁺ (aq) + NO₃⁻(aq)
1 mol 1mol 1 mol
0.001M 0.001M 0.001M
(Complete ionization as HNO₃ is a strong acid)
[HNO₃] = [H⁺] = 0.001 = 1/1000 = 10⁻³ M
pOH = – log [OH⁻]
pOH = – log [10⁻³]
pOH = – (–3) log 10
pOH = 3 × 1
pOH = 3 (∴ log 10 = 1)
pH + pOH = 14
pH = 14 – pOH
pH =14 – 3
pH = 11
Alternate Method
Ionization; HNO₃(aq) ⇌ H⁺ (aq) + NO₃⁻(aq)
1 mol 1mol 1 mol
0.001M 0.001M 0.001M
(Complete ionization as HNO₃ is a strong acid)
[HNO₃] = [H⁺] = 0.001 = 1/1000 = 10⁻³ M
pH = – log [H⁺]
pH = – log [1 × 10⁻³]
pH = – [log 1 + log 10⁻³] (∴ log ab = log a + log b)
pH = – [0 + (–3)] (∴ log 1 = 0 and log 10⁻³ = –3 )
pH = – [0 – 3)]
pH = 3
pH + pOH = 14
pH = 14 – pOH
pH = 14 – 3
pH = 11
Q4. A solution of hydrochloric acid is 0.01 M. What is its pH value?
Solution
Ionization; HCl(aq) ⇌ H⁺ (aq) + Cl⁻(aq)
1 mol 1mol 1 mol
0.01 M 0.01 M 0.01 M
(Complete ionization as HCl is a strong acid)
[HCl] = [H⁺] = 0.01 or 10⁻² M
pH = – log [OH⁻]
pH = – log [10⁻²]
pH = – (–2) log 10
pH = 2 × 1 (∴ log 10 = 1)
pH = 2
Alternate Method
Ionization Equation
Ionization; HCl(aq) ⇌ H⁺ (aq) + Cl⁻(aq)
1 mol 1mol 1 mol
0.01 M 0.01 M 0.01 M
(Complete ionization as HCl is a strong acid)
[HCl] = [H⁺] = 0.01 or 1 × 10⁻² M
pOH = – log [OH⁻]
pOH = – log [1 ×10⁻²]
pOH = – [log 1 + log 10⁻²] (∴ log ab = log a + log b)
pOH = – [0 + (– 2)] ( ∴ log 1 = 0 and log 10⁻² = – 2 )
pOH = – [0 – 2)]
pOH = – [–2]
pOH = 2
pH + pOH = 14
pH = 14 – pOH
pH = 14 – 2
pH = 12
Q6. Find the pH of 0.01 M sulphuric acid.
Solution
Ionization; H₂SO₄(aq) ⇌ 2H⁺ (aq) + SO₄²⁻
1 mol 2mol 1 mol
0.01 M 2×0.01 M 0.01 M
(Complete ionization as H₂SO₄ is a strong acid)
[H₂SO₄] = 2[H⁺] = 2 × 0.01 or 2 × 10⁻² M
pH = – log10 [H⁺]
pH = – log (2 × 10⁻²)
pH = – (–2) log 2
pH = 2 – 0.3
pH = 1.7
Q7. Calculate the pH of a solution whose H⁺ ion concentration is 5 × 10⁻⁴ M.
Solution
Concentration of H⁺ ion = [H⁺] = 5 × 10⁻⁴ M
pH of solution = ?
pH of solution = –log [H⁺]
pH of solution = –log [5 × 10⁻⁴]
pH of solution = –[log 5 + log 10⁻⁴] (∴ log ab = log a + log b)
pH of solution = –[0.6989 + (–4)] (∴log 5 = 0.6989 & log 10⁻⁴=–4)
pH of solution = –[0.6989 – 4]
pH of solution = – [–3.301]
pH of solution = 3.301
Q8. Calculate the pH and pOH of a solution whose H⁺ ion concentration is 3.0 × 10⁻² mol/dm³.
Solution
Concentration of H⁺ ion [H⁺] = 3.0 x 10⁻²M
pH of solution = ?
pOH of solution = ?
pH of solution = – log [H⁺]
pH of solution = – log [3.0 x 10⁻²]
pH of solution = – [log 3.0 + log 10⁻²] (∴ log ab = log a + log b)
pH of solution = – [0.477 + (–2)] (∴ log 3 = 0.477 & log 10⁻² = –2)
pH of solution = – [0.477 – 2]
pH of solution = – [– 1.523]
pH of solution = 1.523
pH + pOH = 14
pOH of solution = 14 – 1.523 = 12.477
Q9. Find out pH and pOH of 0.1 M solution of HCl.
Solution
Concentration of H⁺ ion = [H⁺] = 0.1 M = 10⁻¹ M (as HCl is strong acid)
pH of solution = ?
pH of solution = – log [H⁺]
pH of solution = – log [10⁻¹]
pH of solution = – [log 1 + log 10] ( ∴ log ab = log a + log b)
pH of solution = – [0 + (–1)] ( ∴ log 1 = 0 and log 10⁻¹ = – 1)
pH of solution = – [0–1] = – [–1] = 1
pH + pOH = 14
pOH of solution = 14 – pH
pOH of solution = 14 – 1 = 13
Q10. Find out pOH and pH of 0.01 M solution of NaOH.
Solution
Concentration of OH⁻ ion = [OH⁻] = 0.01 M = 10⁻²M (as NaOH is strong base)
pH of solution = ?
pOH of solution = ?
pOH of solution = – log [OH⁻]
pOH of solution = – log (10⁻²)
pOH of solution = – log (10⁻²)
pOH of solution = – (–2) log 10
pOH of solution = 2 x 1 (∴ log 10 = 1)
pOH of solution = 2
pH + pOH = 14
pH of solution = 14 – pOH
pH of solution = 14 – 2 = 12
🔥📘Assignment Numericals
1. Calculate the pH and pOH of 0.2 M H₂SO₄?
2. Calculate the pH of 0.1 M KOH?
3. Calculate the pOH of 0.004 M HNO₃?
4. Calculate pH of 5M solution of NaOH
5. A solution of H₂SO₄ has pH of 1.05, calculate its pOH and [H⁺]
6. The hydrogen ion concentration of a solution if 1 x 10⁻⁸ mol.dm⁻³, what is the pH of the solution?
7. Complete the following Table.
|
Concentration
of Solution |
[H⁺] |
[OH⁻] |
pH |
pOH |
|
(i) 0.15
M HI |
|
|
|
|
|
(ii) 0.040
M KOH |
|
|
|
|
|
(iii) 0.020
M Ba(OH)₂ |
|
|
|
|
|
(iv) 0.00030
M HClO₄ |
|
|
|
|
|
(v) 0.55
M NaOH |
|
|
|
|
|
(vi) 0.055
M HCl |
|
|
|
|
|
(vii) 0.055
M Ca(OH)₂ |
|
|
|
|
Q1. Calculate the H⁺ ion concentration of a solution whose pH is 4.4?
Solution
pH of solution = – log [H⁺]
log [H⁺] = – pH of solution = – 4.4 (On rearranging the equation)
Since exponent should be the whole number, we add and subtract the whole number that is closest to and larger than the negative log and then Take antilog on both sides
log [H⁺] = – 4.4
log [H⁺] = (– 4.4 + 5) – 5
log [H⁺] = (0.6 – 5)
[H⁺] = antilog (0.6 – 5) = antilog 0.6 × antilog – 5 = 3.98 × 10⁻⁵ = 3.98 × 10⁻⁵ M
Alternate Method of Calculation of [H⁺]
pH = – log [H⁺] ⇒ 10^x = [H⁺] ⇒ 10⁻ᴾᴴ = [H⁺] ⇒ 10⁻⁴˙⁴ = [H⁺] ⇒ [H⁺] = 3.98 × 10⁻⁵ M
Q2. Calculate the H⁺ ion concentration of a solution whose pH = 9.63?
Solution
pH of solution = – log [H⁺]
log [H⁺] = – pH of solution = – 9.63 (On rearranging the equation)
Since exponent should be the whole number, we add and subtract the whole number that is closest to and larger than the negative log then Take antilog on both sides.
log [H⁺] = – 9.63
log [H⁺] = (–9.63 + 10) –10
log [H⁺] = ( 0.37 – 10)
[H⁺] = antilog (0.37 –10)
[H⁺] = antilog 0.37 × antilog –10
[H⁺] = 2.34 × 10⁻¹⁰
[H⁺] = 2.34 × 10⁻¹⁰ M
Alternate Method of Calculation of [H⁺]
pH = – log [H⁺] ⇒ 10^x = [H⁺] ⇒ 10⁻ᴾᴴ = [H⁺] ⇒ 10−9.63 = [H⁺] ⇒ [H⁺] = 2.34 × 10⁻¹⁰M
Q11. Calculate the pH of 0.5 M solution of NaOH.
Solution
Concentration of NaOH = Concentration of OH⁻ ion = [OH⁻] = 0.5 M = 5 × 10⁻¹ M (as NaOH is strong base)
pH of solution = ?
pOH of solution = ?
pOH of solution = – log [OH⁻]
pOH of solution = – log [5 × 10⁻¹]
pOH of solution = –[log 5 + log 10⁻¹] (∴ log ab = log a + log b)
pOH of solution = –[0.698 + (–1)] (∴ log 5 = 0.698 & log 10⁻¹ =–1)
pOH of solution = – [0.698–1]
pOH of solution = – [– 0.302]
pOH of solution = 0.302
pH + pOH = 14
pH of solution = 14 – pOH
pH of solution = 14 – 0.302 = 13.698
Q12. A solution of H₂SO₄ has pH of 1.05. Calculate its pOH and [H⁺].
Solution
Calculation of pOH
pH + pOH = 14 ⇒ pOH = 14 – pH = 14 – 1.05 = 12.95
Calculation of [H⁺]
pH = – log [H⁺]
10^x = [H⁺]
10⁻ᴾᴴ = [H⁺] ⇒ 10⁻¹˙⁰⁵
= [H⁺] ⇒ [H⁺] = 7.08 × 10⁻² M
Q13. The hydrogen ion concentration of a solution is 1× 10⁻⁸ mol.dm⁻³. What is pH of the solution.
Solution
pH = – log [OH⁻] ⇒
pH = – log [10⁻⁸]
pH = – (–8) log 10
pH = 8 × 1 (∴ log 10 = 1)
pH = 8
Q14. Calculate the pH of 0.02 M solution of NaOH. (log 2 = 0.3010)
Solution
Concentration of NaOH = Concentration of OH⁻ ion = [OH⁻] = 0.02 M = 2 × 10⁻² M (as NaOH is strong base)
pH of solution = ?
pOH of solution = ?
pOH of solution = – log [OH⁻]
pOH of solution = – log [2 x 10⁻²]
pOH of solution = – [log 2 + log 10⁻²] (∴ log ab = log a + log b)
pOH of solution = – [0.301 + (– 2)] (∴ log 2 = 0.301 & log 10⁻² = –2)
pOH of solution = – [0.301–2]
pOH of solution = – [– 1.698]
pOH of solution = 1.698
pH + pOH = 14
pH of solution = 14 – pOH
pH of solution = 14 – 1.698 = 12.302
Q15. Calculate the pH and pOH of 0.03 M solution of HCl. (log 3 = 0.4771)
Solution
Concentration of HCl = Concentration of H⁺ = [H⁺] = 0.03 M = 3 × 10⁻² M (as HCl is a strong acid)
pH of solution = ?
pOH of solution = ?
pH of solution = – log [H⁺]
pH of solution = –log [3 × 10⁻²]
pH of solution = –[log 3 + log 10⁻²] (∴ log ab = log a + log b)
pH of solution= –[0.4771 + (– 2)] (∴log 2 =0.4771 & log 10⁻²=–2)
pH of solution= –[0.4771–2]
pH of solution= –[– 1.698] = 1.533
pH + pOH = 14
pOH of solution = 14 – pH
pH of solution = 14 – 1.533 = 12.467
🔶 💥 Master pH, pOH & [H⁺] Concentration with These MDCAT-Style MCQs! 🚀 | Learn Chemistry Like a Pro 🔬 🔶
Welcome to your one-stop guide to mastering pH, pOH, and [H⁺] concentration with MDCAT-style MCQs! Whether you’re preparing for MDCAT, ECAT, or just want to solidify your chemistry knowledge, these MCQs will take your understanding to the next level. Don’t miss out on the explanations at the end that will make everything click! 🧠💡
1. What is the pH of a solution if the hydrogen ion concentration [H⁺] is 1×10⁻⁷?
🟣 A) 7
🟢 B) 0
🟠 C) 1
🔵 D) 14
2. If the pH of a solution is 4, what is the pOH of the solution?
🟣 A) 10
🟢 B) 9
🟠 C) 4
🔵 D) 7
3. What is the concentration of hydroxide ions [OH⁻] in a solution with a pOH of 3?
🟣 A) 1×10⁻³M
🟢 B) 1×10−11 M
🟠 C) 1×10⁻⁷ M
🔵 D) 1×10−14 M
4. Which of the following represents a solution with a pH of 8?
🟣 A) Weak base
🟢 B) Strong acid
🟠 C) Neutral solution
🔵 D) Weak acid
5. If [H⁺] = 4×10⁻⁵ M, what is the pH of the solution?
🟣 A) 5.4
🟢 B) 4.4
🟠 C) 3.5
🔵 D) 7.4
6. The pH of a solution is 2. What is the [OH⁻] concentration?
🟣 A) 1×10⁻¹² M
🟢 B) 1×10⁻² M
🟠 C) 1×10⁻¹⁰ M
🔵 D) 1×10⁻⁴ M
7. The relationship between pH and pOH is given by:
🟣 A) pH + pOH = 14
🟢 B) pH = pOH
🟠 C) pH - pOH = 14
🔵 D) pH + pOH = 7
8. What is the pH of a solution where [H⁺] = 1×10⁻³ M?
🟣 A) 3
🟢 B) 10
🟠 C) 7
🔵 D) 1
9. What is the [H⁺] concentration in a solution with pH 6?
🟣 A) 1×10⁻⁸ M
🟢 B) 1×10⁻⁶ M
🟠 C) 1×10⁻⁴ M
🔵 D) 1×10⁻¹⁰ M
10. A solution has a pOH of 5. What is the pH?
🟣 A) 9
🟢 B) 5
🟠 C) 7
🔵 D) 10
11. What is the pH of a solution if [OH⁻] is 1×10⁻⁴ M?
🟣 A) 4
🟢 B) 10
🟠 C) 7
🔵 D) 6
12. A solution has a pH of 9. What is the [H⁺] concentration?
🟣 A) 1×10⁻⁹ M
🟢 B) 1×10⁻⁵ M
🟠 C) 1×10⁻³ M
🔵 D) 1×10⁻⁷ M
13. Which of the following statements is true for a solution with a pH of 13?
🟣 A) The solution is acidic.
🟢 B) The solution is neutral.
🟠 C) The solution is strongly basic.
🔵 D) The solution is slightly acidic.
14. If the pH of a solution is 11, what is the pOH?
🟣 A) 3
🟢 B) 14
🟠 C) 7
🔵 D) 10
15. What is the pH of a solution with [H⁺] = 5×10⁻⁸ M?
🟣 A) 8.3
🟢 B) 7.5
🟠 C) 7
🔵 D) 6.5
16. The concentration of [OH⁻] in a solution is 1×10⁻⁹ M. What is the pOH of the solution?
🟣 A) 9
🟢 B) 5
🟠 C) 7
🔵 D) 3
17. What is the pH of a solution with a hydrogen ion concentration of 3×10⁻² M?
🟣 A) 1.5
🟢 B) 2
🟠 C) 4
🔵 D) 0.5
18. A solution has a pH of 6. What is its pOH?
🟣 A) 8
🟢 B) 6
🟠 C) 7
🔵 D) 4
19. What is the [OH⁻] concentration in a solution with pH 10?
🟣 A) 1×10⁻⁴ M
🟢 B) 1×10⁻³ M
🟠 C) 1×10⁻¹⁰ M
🔵 D) 1×10⁻¹⁴ M
20. If a solution has a pOH of 2, what is its pH?
🟣 A) 12
🟢 B) 2
🟠 C) 5
🔵 D) 12
🚨 ANSWERS & EXPLANATIONS 🚨
1. Answer: A) pH = 7
Explanation: A neutral solution has a [H⁺] of 1×10⁻⁷ M, which corresponds to a pH of 7.
2. Answer: B) pOH = 9
Explanation: pH + pOH = 14. Since pH = 4, the pOH must be 14−4=9.
3. Answer: A) 1×10⁻³ M
Explanation: If pOH = 3, [OH⁻] = 10⁻³ M (because [OH−] = 10−pOH
4. Answer: D) Weak acid
Explanation: A pH of 8 suggests a slightly basic solution, often from a weak acid or a weak base that doesn't fully dissociate.
5. Answer: B) pH = 4.4
Explanation: pH = -log [H⁺] = -log 4×10⁻⁵ = 4.4.
6. Answer: A) 1×10⁻¹² M
Explanation: pH = 2, so pOH = 14−2=12. The [OH⁻] concentration is 10−12 M.
7. Answer: A) pH + pOH = 14
Explanation: This is a fundamental relationship between pH and pOH for water at 25°C.
8. Answer: A) pH = 3
Explanation: pH = -log [H⁺] = -log 1×10⁻³ = 3.
9. Answer: B) [H⁺] = 1×10⁻⁶ M
Explanation: pH = 6 means [H⁺] = 10⁻⁶ M.
10. Answer: A) pH = 9
Explanation: pH + pOH = 14. Since pOH = 5, pH = 14−5=9.
11. Answer: B) pH = 10
Explanation: If [OH⁻] = 1×10⁻⁴ M, then pOH = 4. Since pH + pOH = 14, pH = 14−4=10
12. Answer: B) [H⁺] = 1×10⁻⁹ M
Explanation: pH = 9 means [H⁺] = 10⁻⁹ M.
13. Answer: C) The solution is strongly basic.
Explanation: A pH of 13 indicates a strongly basic solution. pH values greater than 7 indicate basicity.
14. Answer: A) pOH = 3
Explanation: If pH = 11, then pOH = 14−11=3.
15. Answer: A) pH = 8.3
Explanation: pH = -log [H⁺] = -log 5×10⁻⁸ = 8.3.
16. Answer: A) pOH = 9
Explanation: pOH = -log [OH⁻] = -log 1×10⁻⁹ = 9.
17. Answer: B) pH = 2
Explanation: pH = -log [H⁺] = -log 3×10⁻² = 2.
18. Answer: A) pOH = 8
Explanation: Since pH = 6, pOH = 14−6=8
19. Answer: A) [OH⁻] = 1×10⁻⁴ M
Explanation: pH = 10 means pOH = 14−10 = 4, and [OH⁻] = 10−4 M.
20. Answer: A) pH = 12
Explanation: pH = 14−2=12.
🎉 Don't forget to share this with your friends! Help them prepare for MDCAT with these essential chemistry MCQs. 📚💥
Tags
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