Question

Calculate the pH of the resulting solution if 26.0 mL of 0.260 M HCl(aq) is added to (a) 36.0 mL of 0.260 M NaOH(aq).(b) 16.0 mL of 0.360 M NaOH(aq).

Answer 1

(a) Number of moles of HCl , n = Molarity x volume in L

                                             = 0.260 M x 0.026 L

                                            = 6.76x10^-3 mol

Number of moles of NaOH , n' = Molarity x volume in L

                                             = 0.260 M x 0.036 L

                                             = 9.36x10^-3 mol

HCl + NaOH NaCl + H₂O

From the balanced equation ,

1 mole of HCl reacts with 1 mole of NaOH

6.76x10^-3 mol of HCl reacts with 6.76x10^-3 mol of NaOH

So 9.36x10^-3 mol - 6.76x10^-3 mol = 2.6x10^-3 mol of NaOH left unreated in the resulting 26.0+36.0 = 62.0 mL of solution,which results basic nature of the solution .

So [NaOH] left = number of moles of NaOH left / total volume of the solution

                      = 2.6x10^-3 mol / 0.062 L

                      = 0.042 M

[OH^-] = [NaOH] = 0.042 M

pOH = - log[OH^-] = - log 0.042 = 1.38

pH = 14-pOH = 14-1.38 = 12.62

(b) Number of moles of HCl , n = Molarity x volume in L

                                             = 0.260 M x 0.026 L

                                            = 6.76x10^-3 mol

Number of moles of NaOH , n' = Molarity x volume in L

                                             = 0.360 M x 0.016 L

                                             = 5.76x10^-3 mol

HCl + NaOH NaCl + H₂O

From the balanced equation ,

1 mole of HCl reacts with 1 mole of NaOH

5.76x10^-3 mol of HCl reacts with 5.76x10^-3 mol of NaOH

So 6.76x10^-3 mol - 5.76x10^-3 mol = 1.0x10^-3 mol of HCl left unreated in the resulting 26.0+16.0 = 42.0 mL of solution,which results acidic nature to the solution .

So [HCl] left = number of moles of HCl left / total volume of the solution

                  = 1.0x10^-3 mol / 0.042 L

                 = 0.024 M

[H⁺] = [HCl] = 0.024 M

pH = - log[H⁺]

    = - log 0.024

    = 1.62