Question

calculate the pH of the resulting solution if 28 mL of .28M HCl is added to 18ml of .38M NaOH

Answer 1

Given

V1 = 28 mL = 0.028 L

M1 = 0.28 M (or mol/L)

No. of moles of HCl n1 = V1* M1 = 0.028 L * 0.28 mol/L = 7.84 * 10^-3 mol

V2 = 18 mL = 0.018 L

M2 = 0.38 M (or mol/L)

No. of moles of NaOH n2 = V2* M2 = 0.018 L * 0.38 mol/L = 6.84 * 10^-3 mol

HCl + NaOH ------> NaCl + H₂O

so 6.84 * 10^-3 mol of HCl and 6.84 * 10^-3 mol of NaCl will react and there will be

(7.84 * 10^-3 - 6.84 * 10^-3 ) = 1 * 10^-3 mol of HCl will be remaining which will dissassociate to give

1 * 10^-3 mol of H⁺ ions will be present in the total solution

total volume of solution = 0.028 +0.018 = 0.046 L of solution.

so Concentration of H⁺ ions = 1 * 10^-3 mol /0.046 L = 0.02174 mol/L

pH = -log [H⁺ ] = -log(0.02174) = 1.66 Answer