Question

The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.2 M NaOH will be??(given pKa for CH3COOH =4.74 and log 2=0.301)

Answer 1

"The pH of a solution obtained by mixing 100 ml" of 02. M of $\mathrm{CH}_{3} \mathrm{COOH}$ with 100 ml of 0.2 M of NaOH will be 8.87.

We can calculate this by using the formula: $7+\frac{1}{2} P K a+\frac{1}{2} \log C$

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:

$C=\frac{\text {Total Moles Formed}}{\text {Total Volume}}$

The moles formed for $\mathrm{CH}_{3} \mathrm{COOH}$ and NaOH will be calculated as mass ÷ total volume in L

$=\frac{0.2}{10}=0.02$

We use the equation: $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$

In this case, $\mathrm{CH}_{3} \mathrm{COOH}$ is partially dissociated whereas NaOH is completely dissociated. The total moles dissociated would be 0.02 and the total volume is 200ml.

$C=\frac{0.02}{0.2}=0.1$

Now using the formula to calculate pH, we get $p H=7+\frac{1}{2} P K a+\frac{1}{2} \log C$

$\begin{array}{c}{=7+\frac{1}{2}[P k+\log C]} \\ {=7+\frac{1}{2}[4.74+\log 0.1]} \\ {=7+\frac{1}{2}[4.74+(-1)]}\end{array}$

$\begin{array}{l}{=7+\frac{1}{2}[3.74]} \\ {=7+1.87} \\ {=8.87}\end{array}$

Answer 2

Answer:

Explanation:

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