Question

Question: Consider 50.0 mL of a solution of weak acid HA (Ka = 1.00 ×10^-6), which has a pH of 4.000. What volume of water must be added to make the pH = 5.000?

Short answer

Answer

Water needs to be added = 4950 mL

Step-by-step solution

Calculating Concentration of Acid at pH 4

First of all, lets calculated the value of H⁺ ions

$$\begin{gathered} \text{pH}={\text{- log[H}}^{\text{+}}\text{]} \\ \text{pH}=\text{4}\Rightarrow {\text{[H}}^{\text{+}}\text{]}={\text{10}}^{\text{- pH}}={\text{10}}^{\text{- 4}} \end{gathered}$$

As we know for weak acids

$\left[H^{+}\right]=\sqrt{K_a\times C}$

So, now concentration of acid with pH = 4 can be calculated as

$$\begin{gathered} \left[H^{+}\right]={10}^{-4}=\sqrt{1\times {10}^6\times C} \\ C=\frac{{10}^{-8}}{{10}^{-6}} \end{gathered}$$

= 0.01

Calculating Moles of Acid Present 

Moles of acid can be calculated as

Calculating Concentration of Acid at pH 5 

First, lets calculated the value of H⁺ ions

$\left[H^{+}\right]=\sqrt{K_a\times C}$

As we know for weak acids

So, now concentration of acid with pH = 5 can be calculated as

$$\begin{gathered} \left[H^{+}\right]={10}^{-4}=\sqrt{1\times {10}^6\times C} \\ C=\frac{{10}^{-10}}{{10}^{-6}}={10}^{-4} \end{gathered}$$

Calculating Volume of Water to be Added

Volume of acid can be calculated as

$$\begin{gathered} \text{C}={\text{10}}^{\text{- 4}}\text{mol/L} \\ \text{C}=\frac{\text{Mole}}{\text{Volume(L})}=\frac{5\times {10}^{-4}}{\text{Volume(L})} \\ \text{Volume(L)}=\frac{{\displaystyle 5\times {10}^{-4}}}{{\displaystyle \text{Volume(L})}}=5L=5000ml \end{gathered}$$

So total v,olume of the solution = 5000 mL

Water needs to be added = 5000-50 =4950 mL