Calculate the ph of the resulting solution if 31.0 ml of 0.310 m hcl(aq) is added to

Missing question: 36.0 mL of 0.310 M NaOH(aq).
Chemical reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l).
Ionc net reaction: H⁺(aq) + OH⁻(aq) → H₂O(l)
c(HCl) = 0,310 M = 0,31 mol/L.
V(HCl) = 31,0 mL ÷ 1000 mL/L = 0,031 L.
n(HCl) = c(HCl) · V(HCl).
n(HCl) = n(H⁺) = 0,31 mol/L · 0,031 L.
V(HCl) = 0,00961 mol.
n(NaOH) = 0,036 L · 0,31 mol/L.
n(NaOH) = n(OH⁻) = 0,01116 mol.
n(OH⁻) = 0,01116 mol - 0,00961 mol = 0,00155 mol.
c(OH⁻) = 0,00155 mol ÷ (0,031 L + 0,036 L).
c(OH⁻) = 0,023 M.
pOH = -log(0,023 M) = 1,63.
pH = 14 - 1,63 = 12,37.

Less stable  hope this