Solutions for the “Calculation of pH in buffer systems”

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[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
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Solutions for the “Calculation of pH in buffer systems”

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