calculate the pH of the solution obtained by mixing 100 ml 0.01( M )acetic acid and 20ml 0.01(M) KOH. Given PKa=4.74
Answers
Answer:
8.87
Explanation:
"The pH of a solution obtained by mixing 100 ml" of 02. M of \mathrm{CH}_{3} \mathrm{COOH}CH3COOH with 100 ml of 0.2 M of NaOH will be 8.87.
We can calculate this by using the formula: 7+\frac{1}{2} P K a+\frac{1}{2} \log C7+21PKa+21logC
We first need to calculate the value of C which is formed after the mixture is formed. The value C or final mixture concentration can be calculated as:
C=\frac{\text {Total Moles Formed}}{\text {Total Volume}}C=Total VolumeTotal Moles Formed
The moles formed for \mathrm{CH}_{3} \mathrm{COOH}CH3COOH and NaOH will be calculated as mass ÷ total volume in L
=\frac{0.2}{10}=0.02=100.2=0.02
We use the equation: \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}CH3COOH+NaOH→CH3COONa+H2O
In this case, \mathrm{CH}_{3} \mathrm{COOH}CH3COOH is partially dissociated whereas NaOH is completely dissociated. The total moles dissociated would be 0.02 and the total volume is 200ml.
C=\frac{0.02}{0.2}=0.1C=0.20.02=0.1
Now using the formula to calculate pH, we get p H=7+\frac{1}{2} P K a+\frac{1}{2} \log CpH=7+21PKa+21logC
\begin{gathered}\begin{array}{c}{=7+\frac{1}{2}[P k+\log C]} \\ {=7+\frac{1}{2}[4.74+\log 0.1]} \\ {=7+\frac{1}{2}[4.74+(-1)]}\end{array}\end{gathered}=7+21[Pk+logC]=7+21[4.74+log0.1]=7+21[4.74+(−1)]
\begin{gathered}\begin{array}{l}{=7+\frac{1}{2}[3.74]} \\ {=7+1.87} \\ {=8.87}\end{array}\end{gathered}=7+21[3.74]=7+1.87=8.87