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The $K_a$ of $HA$ is $1.0 \times 10^{-5}$.

How much water should be added to a 100 mL $HA$ solution with a pH of 4 to make the pH = 5?

A

$5.0 \times 10^1 \ \text{mL}$

B

$1.0 \times 10^2 \ \text{mL}$

C

$9.9 \times 10^3 \ \text{mL}$

D

$9.1 \times 10^4 \ \text{mL}$

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