Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
[tex] n_{NaOH}} = n_{i_{NaOH}} - n_{HCl} = 3.60 \cdot 10^{-3} moles - 5.95 \cdot 10^{-4} moles = 3.01 \cdot 10^{-3} moles [/tex]
Now, we need to find the concentration of the OH⁻ ions.
[tex] [OH^{-}] = \frac{n_{NaOH}}{V} [/tex]
Where V is the volume of the solution = 1.00 L
[tex] [OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L [/tex]
Finally, we can calculate the pH of the solution as follows:
[tex] pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52 [/tex]
[tex] pH + pOH = 14 [/tex]
[tex] pH = 14 - pOH = 14 - 2.52 = 11.48 [/tex]
Therefore, the pH of the solution is 11.48.
I hope it helps you!
