To a 100.0 mL volumetric flask were added 7.00 mL of 1.70 M of H3VO4, 1.401...

To a 100.0 mL volumetric flask were added 7.00 mL of 1.70 M of H3VO4, 1.401 g of NaH2VO4, and 15.00 mL of 1.5 M NaOH. The flask was filled to the calibration mark with water. What were the molar concentrations of H+, OH-, the two other relevant molecules and the pH of the solution? Is the resulting solution a buffer? Ka1 = 1.0 x 10-4, Ka2 = 2.8 x 10-9 and Ka3 = 5.0 x 10-15 for H3VO4.

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Expert Solution

moles of H3VO4 in the solution = MxV = 1.70 mol/L x 0.007 L = 0.0119 mol

Moles of NaH2VO4 in the solution = mass / molar mass = 1.401 g / 139.95 g/mol = 0.0100 mol

Moles of NaOH added = MxV = 1.5 mol/L x 0.015 L = 0.0225 mol

Since H3VO3 is a triprotic acid, when NaOH is added will react with 0.0225 / 3 = 0.0075 mol H3VO4 through the following equation

------------ H3VO4 + 3NaOH ----------- > VO₄^3-(aq) + 3H2O(l)

Init.mol: 0.0119, 0.0225 ---------------- 0 mol

change: - 0.0075, (0.0225 - 3x0.0075), +0.0075 mol

eqm.mol: 0.0044, 0 mol, ------------------ 0.0075 mol

Since there are both weak acid H3VO4 and conjugate base H₂VO₄^-(aq), the solution will act as buffer solution. (answer)

Now we can calculate the pH of the buffer solution by applying Hendersen equation

pH = pKa1 + log[H₂VO₄^-(aq)] / [H3VO4] = pKa1 + log(moles of NaH₂VO₄ / moles of H3VO4)

=> pH = - log(1.0x10^-4) + log(0.0100 / 0.0044) = 4.36 (answer)

=> [H+] = 10^-4.36 = 4.4 x 10^-5 M (answer)

=> [OH-] = Kw / Ka = 10^-14 / 4.4 x 10^-5 = 2.27x10^-10 M (answer)

other relevent species are

[VO₄^3-] = 0.0075 mol / 0.100 L = 0.075 M (answr)

queen_honey_blossom answered 11 months ago

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