Question

50ml of 2n acetic acid mixed with 10 ml of1n sodium acetate solution will have an approximate ph of (ka 10^-5 )

Answer 1

According to given data,

Eq.moles of acetic acid = 0.05×2 =  0.1

Eq.moles of sodium acetate = 0.01×1 = 0.01

On substituting above values in Henderson-Hasselbalch equation

$pH = pK_{a} + log\frac{concentration of salt}{concentration of acid}$

$pH = 5 + log\frac{0.01}{0.1} = 5+log(0.1) = 5-1 = 4$

Thus, the pH of resulting solution = 4

Answer 2

Answer: The pH of the solution is 4.

Explanation: As the acid and salt form an acidic buffer of acetic acid. So, to calculate the pH of the solution, we will use Henderson-Hasselbalch equation.

Equation is:

$pH=pK_a+log\frac{\text{concentration of salt}}{\text{concentration of acid}}$

$pK_a=-\log{10^{-5}}=5$

As the acidity of Acetic acid is 1, so normality will be equal to the molarity of the solution.

Moles is calculated by using the formula:

$Moles=Molarity\times Volume$

Moles of acetic acid = 2 × 0.05 = 0.1

Moles of Sodium acetate = 1 × 0.01 = 0.01

Now, the total volume of the solution after mixing is 60mL

So, concentration of acetic acid = $\frac{0.1}{0.06}$

Concentration of Sodium Acetate = $\frac{0.01}{0.06}$

Putting values in Henderson-Hasselbalch equation, we get:

$pH=5+log\frac{0.01}{0.1}$

pH = 4