Question

Consider 1.0 L of a solution which is 0.80 M HF and 0.55 M NaF (K for HF - 7.2x104). Part 1 Calculate the pH of this solution. pH- Part 2 Calculate the pH after 0.10 mol of HCl has been added to the original buffer solution. Assume no volume change on addition of HCI. pH- Part 3 Calculate the pH after 0.20 mol of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH. pH-

Answer 1

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pka = -log ka = -10g 7.2 x10 = 314 [HF] = 0.80M [naf] = 0.55M It forms a buffer solution - PH = pka + log (sall pH = 3-14 +log 090 C PH = [2.98] © molarity of HCI = 0.imol = 0.10M it reacts with Nof. Naf + HCl HF + Nach I(M) 0.55 0.10 0.80 - ccm) - 0.0 -0:10 +0.10 - E(m) 55-6. Dogom 0-45M 10L PH= 3:14 +log 0.45 = 2.841, 0.90 pH decreases on addition of HCl ③ molarity of Naon = 0.20mol = 0.20m it reacts with HE TOOL Ilm) ccm E(m) NaOH + HF Naf +40 0-20 0.20 0.55 -0.20 -0.20 +01-20 - 0.60 0.75 PH= 3,14 +log 0.75 -13.24) 0.60 pH increases on addition of NAOH Scanned with CamScanner