pH Of Acids And Bases MCQ - Practice Questions & Answers

Concepts Covered - 4

pH of Solutions: Strong Acids

pH is also referred to as potential or power of hydrogen. Mathematically, it can be represented as follows:

$\mathrm{pH\: =\: -log_{10}[H_{3}O^{+}]}$

If solution is neutral, then:
K_w = [H₃O⁺][OH^-]
From the ionic product of water, we know:
K_w = 10^-14
[H₃O⁺] = [OH^-] = x (since solution is neutral)
Thus, 10^-14 = K_w = x²
x = 10^-7
Now, [H₃O⁺] = 10^-7
Thus, pH = - log₁₀(H₃O⁺) = - log₁₀(10^-7) = 7

For Acidic solutions: For Basic solutions:
For acidic solutions, we must have [H₃O⁺] > [OH^-] For basic solutions, we must have [H₃O⁺] < [OH^-]
Thus, [H₃O⁺] > 10^-7 Thus, [H₃O⁺] < 10^-7
Thus, [H₃O⁺] for acids can be 10^-6, 10^-5, 10^-4, etc. Thus, [H₃O⁺] for basics can be 10^-8, 10^-9, 10^-10, etc.
Thus, pH of acids can be 6, 5, 4, etc. Thus, pH of basics can be 8, 9, 10, 11, etc.
Hence, pH of acidic solutions is less than 7 Hence, pH of basic solutions is greater than 7

pH depends upon temperature
We know from ionic product of water that at 63⁰C, the value of K_w = 10^-13.
For neutral solution we know:

$\\\mathrm{[H_{3}O^{+}]\: =\: [OH^{-}]}\\\\\mathrm{\Rightarrow\: K_{w}\: =\: x^{2}}\\\\\mathrm{\Rightarrow\: x\: =\: \sqrt{10^{-13}}\: =\: 10^{-6.5}}\\\\\mathrm{\Rightarrow\: [H_{3}O^{+}]\: =\: 10^{-6.5}}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(10^{-6.5})\: =\: 6.5}$

Hence, pH depends upon temperature

pH of Strong Acids
Strong acids are those acids which dissociate completely in solutions. For example:

2 x 10^-3M HNO₃
Since HNO₃ is a strong acid, thus it will dissociate completely into H⁺ and OH^- ions as follows:

$\\\mathrm{HNO_{3}\: \rightarrow \: H^{+}\: +\: NO_{3}^{-}}\\\\\mathrm{Thus,\: [H^{+}]\: =\: 2\: x\: 10^{-3}M\: \: \: \: \: \: \: \: \: (given)}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(2\: x\: 10^{-3})}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(2)\: -\: log_{10}(10^{-3})}\\\\\mathrm{\Rightarrow\: pH\: =\: -0.30\: +\: 3\: =\: 2.7}$
Thus, pH of HNO₃ is 2.7

10^-4 M H₂SO₄
Since H₂SO₄ is a strong acid, thus it will dissociate completely into H⁺ and OH^- ions as follows:

$\\\mathrm{H_{2}SO_{4}\: \rightarrow \: 2H^{+}\: +\: SO_{4}^{2-}}\\\\\mathrm{Thus,\: [H^{+}]\: =\: 2\: x\: 10^{-4}M\: \: \: \: \: \: \: \: \: (given)}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(2\: x\: 10^{-4})}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(2)\: -\: log_{10}(10^{-4})}\\\\\mathrm{\Rightarrow\: pH\: =\: -0.30\: +\: 4\: =\: 3.7}$
Thus, pH of H₂SO₄ is 3.7

NOTE: If molarity(N) of solution is not given but normality(N) is given, then molarity can be calculated using the following formula:

N = M x n
where, n is the number of moles

pH of Solutions: Weak Acids

pH of Weak Acids
Weak acids are those acids which dissociate partially in solutions. For example:

8 M HA (K_a =2 x 10^-8)
The chemical equation for the dissociation of weak acid HA is as follows:
$\mathrm{HA\: \rightleftharpoons \: H^{+}\: +\: A^{-}}$
Initial: 8M 0 0
Equil: 8 - 8𝛂 8𝛂 8𝛂

The equilibrium constant K_a for the weak acid is given as follows:
$\\\mathrm{K_{a}\: =\: \frac{[H^{+}][A^{-}]}{[HA]}\: =\: \frac{8\alpha \, .\, 8\alpha }{8(1-\alpha )}\: =\: \frac{8\alpha ^{2}}{1-\alpha }}\\\\\mathrm{K_{a}\: =\: 8\alpha ^{2}\: \: \: \: \: \: \: (as\: (1-\alpha) \: \approx \: 1)}\\\\\mathrm{Thus,\: \alpha \: =\: \sqrt{\frac{K_{a}}{8}}\: =\: \sqrt{\frac{2\, x\, 10^{-8}}{8}}\: =\: \sqrt{\frac{10^{-8}}{4}}\: =\: 0.5\, x\, 10^{-4}}\\\\\mathrm{[H^{+}]\: =\: 8\,\, x\,\, 0.5\,\, x\,\, 10^{-4}\: =\: 4\: x\: 10^{-4}}\\\\\mathrm{pH\: =\: -log_{10}4\: +\: 4}\\\\\mathrm{pH\: =\: -0.60\: +\: 4\: =\: 3.4}$
Thus, pH of this given acid = 3.4
0.002N CH₃COOH(𝛂 = 0.02)
The chemical equation for the dissociation of CH₃COOH is as follows:
$\mathrm{CH_{3}COOH\: \rightleftharpoons \: CH_{3}COO^{-}\: +\: H^{+}}$
Initial: c 0 0
Equil: c - c𝛂 c𝛂 c𝛂

The equilibrium constant K_a for the weak acid is given as follows:
$\\\mathrm{K_{a}\: =\: \frac{[CH_{3}COOH^{-}][H^{+}]}{[CH_{3}COOH]}\: =\: \frac{c\alpha \, .\, c\alpha }{c(1-\alpha )}\: =\: \frac{c\alpha ^{2}}{1-\alpha }}\\\\\mathrm{K_{a}\: =\: c\alpha ^{2}\: \: \: \: \: \: \: (as\: (1-\alpha) \: \approx \: 1)}\\\\\mathrm{Now,\: as\: we\: have\: given:}\\\mathrm{c\: =\: 0.002N\: or\: 0.002M\: \: \: \: \: \: (Normality\: =\: Molarity,\: as\: n\: factor\: =\: 1)}\\\mathrm{\alpha \: =\:\frac{2}{100}\: =\: 0.02}\\\\\mathrm{Thus,\: [H^{+}]\: =\: 0.002\,\, x\,\, 0.02\: =\: 4\: x\: 10^{-5}}\\\\\mathrm{pH\: =\: -log_{10}(4\: x\: 10^{-5})}\\\\\mathrm{pH\: =\: 5\: -\:log 4\: =\: 4.4}$
Thus, pH of acetic acid = 4.4

pH of Solutions: Strong Bases

Strong bases
Strong bases are those bases which dissociate completely in solution. For exaple:

0.1M NaOH
Since NaOH is a strong base, thus it will dissociate completely into Na⁺ and OH^- ions. The chemical equation for the dissociation of NaOH is as follows:
$\\\mathrm{NaOH\: \rightarrow Na^{+}\: +\: OH^{-}}\\\\\mathrm{[OH^{-}]\: =\: 10^{-1}M}\\\\\mathrm{Thus,\: pOH\: =\: -log_{10}(OH^{-})\: =\: -log_{10}(10)^{-1}\: =\: 1\: \: \: \: \: \: \: .................(1)}\\\\\mathrm{For\: any\: solution,\: we\: know:}\\\mathrm{pH\: +\: pOH\: =\: 14}\\\\\mathrm{Thus,\: from\: equation(1),\: we\: have:}\\\mathrm{pH\: =\: 13}$
Hence, pH of 0.1M NaOH solution is 13.
0.05M Ba(OH)₂
Since Ba(OH)₂ is a strong base, thus it will dissociate completely into Ba²⁺ and 2OH^- ions. The chemical equation for the dissociation of Ba(OH)₂ is as follows:
$\\\mathrm{Ba(OH)_{2}\: \rightarrow Ba^{2+}\: +\: 2OH^{-}}\\\\\mathrm{[OH^{-}]\: =\: 2\: x\: 5\: x\: 10^{-2}\: M\: =\: 10\: x\: 10^{-2}\: =\: 10^{-1}}\\\\\mathrm{Thus,\: pOH\: =\: -log_{10}(OH^{-})\: =\: -log_{10}(10)^{-1}\: =\: 1\: \: \: \: \: \: \: .................(1)}\\\\\mathrm{For\: any\: solution,\: we\: know:}\\\mathrm{pH\: +\: pOH\: =\: 14}\\\\\mathrm{Thus,\: from\: equation(1),\: we\: have:}\\\mathrm{pH\: =\: 13}$
Hence, pH of 0.05M Ba(OH)₂ solution is 13.

pH of solution/mixture

Mixture of Strong Acids:

$\mathrm{\mathbf{10^{-2}M\: HCl(1L)+10^{-2} M\: HNO_{3}(2 L)}}\\\\\mathrm{}$

$\\\mathrm{The\: chemical \: equation\: for \: HCl\: is\: as\: follows:}\\\mathrm{HCl\: \rightarrow H^{+}\: +\: Cl^{-}}\\\mathrm{Thus,\: moles\: of\: H^{+}\: =\: 10^{-2}\, M}\\\\\mathrm{Similarly,\: for\: HNO_{3},\: we\: have:}\\\mathrm{HNO_{3}^{-}\: \rightarrow \: H^{+}\: +\: NO_{3}^{-}}\\\mathrm{Thus,\: moles\: of\: H^{+} =\: 2\: x\: 10^{-2}\, M\: \: \: \: \: \: \: \: \: \: (since\: volume\: is\: 2L)}\\\\\mathrm{Thus,\: total\: moles\: of\: H^{+}=\: 3\: x\: 10^{-2}\: and\: total\: volume\: =\: 3L}\\\mathrm{Thus,\: [H^{+}]\: =\: \frac{3\: x\: 10^{-2}}{3}\: =\: 10^{-2}\, M}\\\\\mathrm{\mathbf{pH\: =\: -log_{10}(10^{-2})\: =\: 2}}$

NOTE: Shortcut only for monobasic acids and monoacidic bases:

$\\\mathrm{[H^{+}]\: =\: \frac{M_{1}V_{1}\: +\: M_{2}V_{2}}{V_{1}\: +\: V_{2}}\: \: \: \: \: \: (for\: acids)}\\\\\mathrm{[OH^{-}]\: =\: \frac{M_{1}V_{1}\: +\: M_{2}V_{2}}{V_{1}\: +\: V_{2}}\: \: \: \: \: \: (for\: bases)}$
Mixture of Strong Bases:

$\mathrm{\mathbf{10^{-3} M\: NaOH(2L)+10^{-2} M\: KOH(1L)}}$
Using the shortcut formula for bases as given above, we get:
$\\\mathrm{\left[OH^{-}\right]=\frac{M_{1} V_{1}+M_{2} V_{2}}{V_{1}+V_{2}}=\frac{10^{-3} \times 2+10^{-2} \times 1}{3}}\\\\\mathrm{\Rightarrow\: \frac{10^{-3}(2+10)}{3}=\frac{12 \times 10^{-3}}{3}=4 \times 10^{-3}}\\\\\mathrm{pOH=-\log 4+3=-0.60+3=2.40}\\\\\mathrm{Thus,\: p H=14-p O H=14-2.40=11.60}$
Mixture of Strong Acid and Strong Base:

$\mathrm{\mathbf{10^{-2}M\, HCl(2L)+10^{-3} M\, NaOH(1L)}}$
Clearly, moles of (H⁺) = 2 x 10^-2moles and moles of (OH^-) = 1 x 10^-3 moles.
Since moles of (H⁺) is greater than moles of (OH^-), therefore the solution medium will be acidic.
Now, remaining moles of H⁺ = 2 x 10^-2 - 10^-3 = 19 x 10^-3
$\\\mathrm{Thus,\: [H^{+}]\: =\: \frac{19\: x\: 10^{-3}}{3}\: =\: 6.3\: x\: 10^{-3}}\\\\\mathrm{Thus,\: pH\: of\: mixture\: =\: -log_{10}(6.3\: x\: 10^{-3})}\\\\\mathrm{pH\: =\: 3\: -\:log_{10}6.3\: =\: 3\: -\: 0.79\: =\: 2.2}$
Thus, pH of the mixture = 2.2