Solutions for the “Calculation of pH in buffer systems”

Solutions for the “Calculation of pH in buffer systems”1. What is the pH of the buffer solution in which the concentration of acetic acid is0.200 M and that of sodium acetate is 0.500 M?c acetic acid = 0.200 Mc Na-acetate = 0.500 MK a = 1.86 × 10 –5[H + cweak acid] = K acweak basepH = – log [H + ]=1.86 × 10− 5 ×0.2000.500= 7.44 × 10 –6 MpH = – log 7.44 × 10 –6pH = 5.1282. What is the pH of the buffer solution that was prepared by mixing 100 mL of 0.200 MNH 3 solution and 40.00 mL of 0.100 M HCl solution?V(NH 3 solution) = 100 mLn(base) = n base - n HClc(NH 3 solution) = 0.200 M n(base) = 0.02 – 0.004n(NH 3 ) = 0.1 × 0.2 = 0.02 moln(base) = 0.016 molV(HCl solution) = 40 mLc(HCl solution) = 0.100 Mn(HCl) = 0.1 × 0.04 = 0.004 moln(salt) = 0.004 mol[OH – n] = K bnweak basesalt=1.75×10− 5 ×0.0160.004= 7 × 10 –5 MpOH = – log [OH – ]pOH = – log 7 × 10 –5pOH = 4.155 pH = 14.000 – 4.155 = 9.8453. What is the buffer capacity of an acidic buffer in which c salt = c acid = 1.0 M? What isthe change of pH in this buffer upon 10-fold dilution?[H + nweak acid1] starting = K a = K a = Kanweak base 1The effect of a strong acid (molar amount added to 1 L buffer is Y):1

[H + ] = 10×[H + 1+ Y] starting = 10 × K a = K a ×1- Y1+ Y10 =1- Y10 – 10Y = 1 + YY = 0.81818 MThe effect of a strong base (molar amount added to 1 L buffer is Z):[H + ] = 0.1×[H + 1 − Z] starting = 0.1 × K a = K a ×1 + Z1 − Z0.1 =1 + Z0.1 + 0.1Z = 1 – ZZ = 0.81818 MDilution does not change the pH of this buffer, because [H + n] = K annweak acidratio does not cange with dilution.nweak baseweak acidweak baseand the4. What is the buffer capacity of a 1.0 M HCl solution? What is the change of pH in thisbuffer upon 10-fold dilution?[H + ] starting = 1 M, so pH starting = 0.00The effect of a strong acid:[H + ] = 10 MThe added amount of strong acid is 10M – 1M = 9 MThe effect of a strong base:[H + ] = 0.10 MThe added amount of strong base is 1M – 0.1M = 0.9 MAfter a 10-fold dilution, [H + ] = 0.10 M, pH = 1.005. What volume of 0.1 M NaOH or 0.1 M HCl causes 1 unit change in the pH of 1 dm 3 ofacetate buffer, which has a pH of 5.000 and the sum of the acetate and the acetic acidconcentrations is exactly 1.00 M?pH = 5.00, so [H + ] = 10 -5 M[H + nweak acid] = K anweak base10 -5 = 1.86×10 -5 nweak×nacidweak base (orsalt)= 1.86×10 -5 n×1- nweak acidweak acidBy solving this equation, n weak acid = 0.34965 M and n weak base = 0.65035 MThe effect of a strong acid (molar amount: X):pH = 5.00 – 1 = 4.00, so [H + ] = 10 -4 M10 -4 = 1.86×10 -5 0.34965 + X×0.65035 - XX = 0.49352 mol strong acid can be added to decrease the pH by 1 unit.We know the concentration (0.1 M), so the volume can be calculated:2

n 0.49352 molV = == 4.9352 Lc 0.1 mol/LThe effect of a strong base (molar amount: Y):pH = 5.00 + 1 = 6.00, so [H + ] = 10 -6 M10 -6 = 1.86×10 -5 0.34965 − Y×0.65035 + YY = 0.2986 mol strong base can be added to increase the pH by 1 unit.We know the concentration (0.1 M), so the volume can be calculated:n 0.2986 molV = == 2.986 Lc 0.1 mol/L6. A buffer with pH = 9.25 is needed from 100 cm 3 of 2.00 M ammonia solution and anunlimited amount of solid NH 4 Cl. What is the mass of ammonium chloride needed?What is the final pH of this buffer solution after the addition of 50 or 500 cm 3 of 1.00M HCl?pH = 9.25pOH = 14 – pH = 4.75[OH - ] = 10 -pOH = 1.778×10 -5 M[OH - cweak base] = K b ⋅csalt1.778 × 10 -5 = 1.75 × 10 -5 2 M⋅csaltFrom this, c salt = 1.9682 M. We know the volume (100 mL), so the molar amount of thesalt can be calculated: n = c × V = 1.9682 M × 0.1 L = 0.19682 mol.We also know the M.W. for NH 4 Cl (53.5 g/mol), so m = n × M.W. = 10.53 g NH 4 ClAfter the addition of 50 mL 1 M HCl:n HCl = 0.050 mL × 1 M = 0.050 mol[OH - nweak base] = K b ⋅ = 1.75 × 10 -5 0.2 - 0.050⋅=1.06352 × 10 -5 Mnsalt0.19682 + 0.050pOH = -log(1.06352 × 10 -5 ) = 4.973, pH = 14.00 – pOH = 9.027After the addition of 500 mL 1 M HCl:n HCl = 0.50 mL × 1 M = 0.50 molThe total volume of the resulting solution is 600 mL (500 mL + 100 mL)In this solution, we have 0.39682 mol NH 4 Cl (0.2 + 0.19682) and 0.3 mol HCl (0.5 – 0.2).So, take only the strong acid into consideration.0.3 molc HCl = = 0.5 M = [H + ], and pH = -log(0.5) = 0.3010.6 L3