Question

You are given 100 mL of a 0.1 M solution of HCl and 200 mL of a 0.2 M solution of NaOH. Calculate the pH of the resulting solution after mixing the two solutions.

• A. pH = 7
• B. pH = 9
• C. pH = 10
• D. pH = 11

Hint:

When mixing acid and base solutions, first neutralize the excess acid or base, then calculate the remaining concentration to determine pH.

Answer 1

The Correct Option is C

First, calculate the moles of HCl and NaOH: \[ \text{Moles of HCl} = 0.1 \times 0.1 = 0.01 \, \text{mol} \] \[ \text{Moles of NaOH} = 0.2 \times 0.2 = 0.04 \, \text{mol} \] The moles of NaOH exceed those of HCl, so after neutralization, NaOH remains: \[ \text{Excess moles of NaOH} = 0.04 - 0.01 = 0.03 \, \text{mol} \] Now, calculate the concentration of excess NaOH in the final volume (300 mL): \[ \text{Concentration of NaOH} = \frac{0.03}{0.3} = 0.1 \, \text{M} \] Finally, calculate the pOH: \[ \text{pOH} = -\log(0.1) = 1 \] Then, calculate the pH: \[ \text{pH} = 14 - \text{pOH} = 14 - 1 = 10 \]