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Q.

When NaOH solution is gradually added to the solution of a weak acid (HA), the pH of the solution is found to be 5.0 at the addition of 10.0 mL of NaOH and 6.0 at the further addition of 10.0 mL of same NaOH. (Total volume of NaOH=20 mL) calculate ${pK}_{a}$ for HA [log 2=0.3] (Give your answer in nearest integer)

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answer is 5.

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Detailed Solution

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Let initial conc of HA & NaOH be $C_{1}$ & $C_{2}$ mol/L and initial volume of
${HA=V}_{1} mL$
${5.0=pK}_{a} +log \frac{{10C}_{2}}{(C_{1} V_{1} {-10C}_{2})} ....... (1)$

${6.0=pK}_{a} + \log \frac{20 C_{2}}{(C_{1} V_{1} - 20 C_{2})} ...... (2)$
from these we get
$C_{1} V_{1} = 22.5 C_{2}$
$5.0 = p K_{a} + \log \frac{10 C_{2}}{22.5 C_{2} - 10 C_{2}} = p K_{a} + \log (0.8)$
$p K_{a} = 5.1$

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