Question

Question: A 50.00-mL solution of a weak acid HA (K_a = 5.00 × 10^-10) in water has a pH = 5.650. Calculate the amount of water that must be added to reach a pH value of 6.650.

Short answer

Answer

Water needs to be added = 4950 mL

Step-by-step solution

Calculating Concentration of Acid at pH 5.65

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

So, now concentration of acid with pH = 5.65 can be calculated as

Calculating Moles of Acid Present

Moles of acid can be calculated as

$$\begin{gathered} \text{C = 0.01Mol/L} \\ \text{C =}\frac{\text{Mole}}{\text{Volume(L)}}\text{=}\frac{\text{Mole}}{\text{0.05}} \\ {\text{Mole=0.01×0.05=10}}^{-4} \end{gathered}$$

 Step 3: Calculating Concentration of Acid at pH 6.65

Let’s calculate the value of H⁺ ions

So, now concentration of acid with pH = 6.65 can be calculated as

Calculating Volume of Water to be Added

Volume of acid can be calculated as

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

So total volume of the solution = 5000 mL

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