Question

Calculate the pH after 0.10 mole of NaOH is added to 1.00 L of the solution in Exercise 27, and calculate the pH after 0.20 mole of HCl is added to 1.00 L of the solution in Exercise 27.

Short answer

The pH of 0.6 M and 1 M HF and KF buffer solutionafter 0.10 mole of NaOH is added to 1.00 L of the solution and 0.2 moles of HCl to 2 L of solution are3.48 and 3.14respectively.

Step-by-step solution

Given Information

The concentration of HF and KF are 0.6 and 1 M.

Substituting in the equation

It can be calculated as,

$\begin{array}{l}pH=p{K}_a\left(HF\right)+\log \frac{\left[KF\right]}{\left[HF\right]}\\ K_a\left(HF\right)=7.2\times {10}^{-4}\\ pH=-\log \left(K_a\right)+\log \frac{\left[KF\right]}{\left[HF\right]}\\ pH=-\log (7.2\times {10}^{-4})+\log \frac{\left[1M\right]}{\left[0.6M\right]}\\ pH=3.143+0.222\\ pH=3.37\end{array}$

Calculating the pH in addition to 0.1 mole NaOH to 1 L buffer 

It can be denoted as,

$\begin{array}{l}pH=-\log (7.2\times {10}^{-4})+\log \frac{\left[1.1M\right]}{\left[0.5M\right]}\\ pH=3.143+0.342\\ pH=3.48\end{array}$

Calculating the pH in addition to 0.2 mole HCl to 1 L buffer 

It can be denoted as,

$\begin{array}{l}pH=-\log (7.2\times {10}^{-4})+\log \frac{\left[0.8M\right]}{\left[0.8M\right]}\\ pH=3.143+0\\ pH=3.14\end{array}$

The pH of 0.6 M and 1 M HF and KF buffer solutionafter 0.10 mole of NaOH is added to 1.00 L of the solution and 0.2 moles of HCl to 2 L of solution are3.48 and 3.14respectively.