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pH Of Acids And Bases MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • pH of Solutions: Strong Acids, pH of Solutions: Weak Acids, pH of solution/mixture are considered the most difficult concepts.

  • pH of Solutions: Strong Bases are considered the most asked concepts.

  • 17 Questions around this concept.

Solve by difficulty

Which one of the following statements is not true?

Among the following acids which has the lowest pK_{a} value?

In a buffer solution containing an equal concentration of B^{-} and HB, the K_b \ for \ B^{-} is 10^{-10} . The pH of the buffer solution is:

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Which of the following salts will give highest pH in water?

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:
Kw = [H3O+][OH-]
From the ionic product of water, we know:
Kw = 10-14
[H3O+] = [OH-] = x (since solution is neutral)
Thus, 10-14 = Kw = 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 Kw = 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 10-3 M 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 (Ka =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 - 8?         8?         8?

    The equilibrium constant Ka 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(? = 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 - c?                         c?                c?

    The equilibrium constant Ka 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 10-2 moles 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

Study it with Videos

pH of Solutions: Strong Acids
pH of Solutions: Weak Acids
pH of Solutions: Strong Bases

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Reference Books

pH of Solutions: Strong Acids

Chemistry Part II Textbook for Class XI

Page No. : 217

Line : 35

pH of Solutions: Weak Acids

Chemistry Part I Textbook for Class XI

Page No. : 217

Line : 35

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