Calculate the pH of the resulting solution if 16.0 mL. of 0.160 M HCl(aq) is added to 21.0 mL of 0.160 M NaOH(aq

1 answer

To calculate the pH of the resulting solution when 16.0 mL of 0.160 M HCl is mixed with 21.0 mL of 0.160 M NaOH, we can follow these steps:

  1. Calculate the moles of HCl and NaOH:

    • Moles of HCl: \[ \text{Moles of HCl} = \text{Volume (L)} \times \text{Concentration (M)} = 0.016 , \text{L} \times 0.160 , \text{M} = 0.00256 , \text{moles} \]

    • Moles of NaOH: \[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.021 , \text{L} \times 0.160 , \text{M} = 0.00336 , \text{moles} \]

  2. Determine which reactant is in excess:

    • The balanced reaction for HCl and NaOH is: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

    Since both HCl and NaOH react in a 1:1 molar ratio, we can compare their moles:

    • \( \text{Moles of NaOH} = 0.00336 \)
    • \( \text{Moles of HCl} = 0.00256 \)

    Since \( 0.00336 > 0.00256 \), NaOH is in excess.

  3. Calculate the moles of excess NaOH: \[ \text{Excess NaOH} = \text{Moles of NaOH} - \text{Moles of HCl} = 0.00336 - 0.00256 = 0.00080 , \text{moles of NaOH} \]

  4. Calculate the total volume of the solution: \[ \text{Total Volume} = 16.0 , \text{mL} + 21.0 , \text{mL} = 37.0 , \text{mL} = 0.037 , \text{L} \]

  5. Calculate the concentration of excess NaOH: \[ \text{Concentration of excess NaOH} = \frac{\text{Moles of excess NaOH}}{\text{Total Volume (L)}} = \frac{0.00080 , \text{moles}}{0.037 , \text{L}} \approx 0.0216 , \text{M} \]

  6. Calculate the pOH and then the pH: \[ \text{pOH} = -\log[\text{OH}^-] = -\log(0.0216) \approx 1.666 \] \[ \text{pH} = 14 - \text{pOH} \approx 14 - 1.666 \approx 12.334 \]

Thus, the pH of the resulting solution is approximately 12.33.