Calculate the pH of the resulting solution if 16.0 ml of 0.160 M HCl(aq) is added to 26.0 ml. of 0.210 MNaOH

1 answer

To calculate the pH of the resulting solution after mixing HCl and NaOH, we first need to determine the number of moles of HCl and NaOH involved in the reaction.

  1. Calculate the moles of HCl: \[ \text{Moles of HCl} = \text{Concentration (M)} \times \text{Volume (L)} \] \[ = 0.160 , \text{M} \times 0.0160 , \text{L} = 0.00256 , \text{moles of HCl} \]

  2. Calculate the moles of NaOH: \[ \text{Moles of NaOH} = \text{Concentration (M)} \times \text{Volume (L)} \] \[ = 0.210 , \text{M} \times 0.0260 , \text{L} = 0.00546 , \text{moles of NaOH} \]

  3. Determine the reaction: HCl and NaOH react in a 1:1 molar ratio: \[ \text{HCl (aq) + NaOH (aq) } \rightarrow \text{ NaCl (aq) + H}_2\text{O (l)} \]

  4. Calculate the limiting reagent: We have:

    • Moles of HCl = 0.00256
    • Moles of NaOH = 0.00546

    Since HCl is the limiting reagent, all of it will react with NaOH.

  5. Calculate the remaining moles of NaOH: \[ \text{Remaining moles of NaOH} = 0.00546 , - , 0.00256 = 0.00290 , \text{moles} \]

  6. Calculate the total volume of the solution: \[ \text{Total Volume} = 16.0 , \text{ml} + 26.0 , \text{ml} = 42.0 , \text{ml} = 0.0420 , \text{L} \]

  7. Calculate the concentration of the remaining NaOH: \[ \text{Concentration of NaOH} = \frac{\text{remaining moles}}{\text{total volume (L)}} \] \[ = \frac{0.00290}{0.0420} \approx 0.0690 , \text{M} \]

  8. Calculate the pOH of the solution: \[ \text{pOH} = -\log[\text{NaOH}] \] \[ = -\log(0.0690) \approx 1.16 \]

  9. Calculate the pH: \[ \text{pH} + \text{pOH} = 14 \] \[ \text{pH} = 14 - 1.16 \approx 12.84 \]

Thus, the pH of the resulting solution is approximately 12.84.