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  • Calculate the pH of 50 mL of 0.05 M NaOH + 50 mL of 0.1 M CH3COOH. (Given Ka= 1.8 &nbs

Calculate the pH of 50 mL of 0.05 M NaOH + 50 mL of 0.1 M CH3COOH.
(Given Ka= 1.8 \times 10-5 and Kb = 1.8 \times 10-5, pKa = pKb = 4.7447)

Option: 1

4.7447


Option: 2

9.2553


Option: 3

7.2747


Option: 4

2.227


Answers (1)

As we have learnt,

 

Calculating pH of a Buffer Solution(acidic) -

Some examples

  • Find the pH of a solution having 0.1M CH3COOH(Ka = 10-5) and 0.2M CH3COONa.

    We know that pH of a solution is given as:

    \\\mathrm{pH\: =\: pK_{a}\: +\: log_{10}\frac{[Salt]}{Acid}}\\\\\mathrm{Thus,\: pH\: =\: -log_{10}K_{a}\: +\: log_{10}\frac{[0.2]}{[0.1]}}\\\\\mathrm{\Rightarrow\: pH\: = -log_{10}10^{-5}\: +\: log_{10}2}\\\\\mathrm{\Rightarrow\: pH\: =\: 5\: +\: 0.30\: =\: 5.30}
     
  • Find the pH of a solution containing 0.25 moles of HCN(Ka = 10-5) and 0.10 moles of NaCN present in 1 litre solution.

     We know that pH of a solution is given as:
    \\\mathrm{pH\: =\: pK_{a}\: +\: log_{10}\frac{[Salt]}{Acid}}\\\\\mathrm{Thus,\: pH\: =\: -log_{10}K_{a}\: +\: log_{10}\frac{[Salt]}{[Acid]}}\\\\\mathrm{\Rightarrow\: pH\: = -log_{10}10^{-5}\: +\: log_{10}\frac{0.10}{0.25}}\\\\\mathrm{\Rightarrow\: pH\: = 5\: +\: log_{10}\frac{2}{5}}\\\\\mathrm{\Rightarrow\: pH\: =\: 5\: -\: 0.39\: =\: 4.6}

-

\begin{array}{l}{[\mathrm{NaOH}]=2.5 \mathrm{m\: moles} ;[\mathrm{HAc}]=5 \mathrm{m} \text { moles; }\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \\ {\text =2.5 \mathrm{m\: moles} ;[\text { salt }] \text { formed }=2.5 \mathrm{m\: moles}} \\ {\mathrm{pH}=\mathrm{pK}_{a}+\log (\text { salt/acid })=4.7447+\log (2.5 / 2.5)} \\ {=4.7447}\end{array}

Therefore,option(1) is correct

Posted by

Ramraj Saini

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