Question

calculate the ph of the following mixtures given ( k_{a}=1.8\times10^{-5} ) and ( k_{b}=1.8\times10^{-5} ) (( pk_{a}=pk_{b}=4.7447 )).50 ml 0.05 m naoh + 50 ml of 0.1 m ( ch_{3}cooh ) a) 4.7447 b) 8.7447 c) 6.7447 d) 2.7447

Explanation:

Step1: Calculate the moles of reactants

Moles of \(NaOH=n = C\times V\), where \(C = 0.05\space M\) and \(V=50\times10^{- 3}\space L\). So \(n_{NaOH}=0.05\times50\times10^{-3}=2.5\times10^{-3}\space mol\)
Moles of \(CH_{3}COOH=n = C\times V\), where \(C = 0.1\space M\) and \(V = 50\times10^{-3}\space L\). So \(n_{CH_{3}COOH}=0.1\times50\times10^{-3}=5\times10^{-3}\space mol\)

Step2: Determine the reaction and remaining species

The reaction is \(CH_{3}COOH+NaOH
ightarrow CH_{3}COONa + H_{2}O\)
After reaction, moles of \(CH_{3}COOH\) remaining \(n_{acid}=5\times10^{-3}-2.5\times10^{-3}=2.5\times10^{-3}\space mol\)
Moles of \(CH_{3}COONa\) formed \(n_{salt}=2.5\times10^{-3}\space mol\)
Total volume \(V = 50 + 50=100\times10^{-3}\space L\)
Concentration of \(CH_{3}COOH,C_{acid}=\frac{2.5\times10^{-3}}{100\times10^{-3}}=0.025\space M\)
Concentration of \(CH_{3}COONa,C_{salt}=\frac{2.5\times10^{-3}}{100\times10^{-3}}=0.025\space M\)

Step3: Use the Henderson - Hasselbalch equation

The Henderson - Hasselbalch equation for a buffer (weak acid + its salt) is \(pH=pK_{a}+\log\frac{[salt]}{[acid]}\)
Since \(\frac{[salt]}{[acid]}=\frac{0.025}{0.025} = 1\) and \(pK_{a}=4.7447\)
\(pH=4.7447+\log(1)\)
Since \(\log(1) = 0\)

Answer:

A. 4.7447