Calculate the final pH of a solution made by the addition of 10 ml of 0.50 M NaO

Question

Calculate the final pH of a solution made by the addition of 10 ml of 0.50 M NaOH to 10 ml of a
0.55 M solution of a weak acid with a pKa of 5.1.

Explanation / Answer

HOCl + H2O H3O^+ + OCl^- Ka = ([H3O^+][OCl^-])/[HOCl] = 3.5 x 10^-8 So: (Ka x [HOCl]) / [OCl-] = [H3O^+] The reaction between HOCl and NaOH is HOCl + NaOH -> H2O + Na^+ + OCl^- We see that for every mole of HOCl that reacts with NaOH, the number of moles of HOCl will decreases and the number of moles of OCl^- increases. We can work out the initial numbers of moles of HOCl and OCl^- from the equation: Conc. = (number of moles) / Volume For HOCl we have (0.50 M) x (100 x 10^-3 L) = 0.05 moles HOCl For OCl^- we have (0.40 M) x (100 x 10^-3 L) = 0.04 moles OCl^- We now calculate the number of moles of NaOH added: (1.0 M) x (10 x 10^-3 L) = 0.01 moles NaOH We know from the reaction that the number of moles of HOCl will decrease after addition of NaOH and that the number of moles of OCl^- will increase after the addition of NaOH. The number of moles will increase or decrease by 0.01 moles (The number of moles of NaOH added) Now we calculate the number of moles of HOCl and OCl- after the addition of NaOH For HOCl (Initial moles HOCl) - (Moles NaOH added) = Final number of moles of HOCl = 0.05 - 0.01 = 0.04 moles HOCl For OCl^- (Initial moles OCl^-) + (Moles NaOH added= Final number of moles OCl^- = 0.04 + 0.01 = 0.05 moles OCl^- To calculate the final concentrations of HOCl and OCl^- we divide the number of moles by the total volume after the addition of NaOH = 100 mL + 10 mL =110 mL = 0.110 L So [HOCl] = 0.364 M [OCl^-] = 0.454 M Substituting these values in the expression (Ka x [HOCl]) / [OCl-] = [H3O^+] We can calculate [H3O^+] = (3.5 x 10^-8 x 0.364) / 0.454 = 2.8 x 10^-8 M pH = -log[H3O^+] = 7.55

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