# Buffer calculations. 0.842 g of propanoic acid is dissolved in water and mixed with 25.0 mL of 0.108 mol L –1 NaOH. The final solution is made up to 100.0.

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Buffer calculations

0.842 g of propanoic acid is dissolved in water and mixed with 25.0 mL of 0.108 mol L –1 NaOH. The final solution is made up to 100.0 mL with distilled water. M(CH 3 CH 2 COOH) = 74.1 g mol –1, pKa = 4.87 a Calculate the pH of the solution. b Calculate the new pH if 1.00 mL of 1.00 mol L–1 NaOH is added to the solution made above. Find the amount, in moles, of propanoic acid and NaOH added to the mixture.

Some of the acid reacts with the base, according to the following equation: CH 3 CH 2 COOH + NaOH → CH 3 CH 2 COO – + Na + + H 2 O n(CH 3 CH 2 COO – ) = n(NaOH) = 2.70  10 –3 mol n(CH 3 CH 2 COOH) = n(CH 3 CH 2 COOH) start – n(NaOH) = 0.01136 mol – 2.70  10 –3 mol = 8.66  10 –3 mol

To calculate the pH we will need to derive the buffer formula: CH 3 CH 2 COOH + H 2 O CH 3 CH 2 COO – + H 3 O +

To convert the amounts (in moles) of acid and base present in the mixture into concentrations, we divide by the volume of liquid present. The volumes cancel — so we can ignore them and just work with the amounts of acid and base present.

Now to find the pH when 1.00 mL of 1.00 mol L –1 NaOH added. n(NaOH) added = cV = 1.00 mol L –1  1.00  10 –3 L = 1.00  10 –3 mol The NaOH will convert CH 3 CH 2 COOH into CH 3 CH 2 COO – : n(CH 3 CH 2 COO – ) = 2.70  10 –3 mol + 1.00  10 –3 mol = 3.70  10 –3 mol n(CH 3 CH 2 COOH) = 8.66  10 –3 mol – 1.00  10 –3 mol = 7.66  10 –3 mol

The volume of the solution is now 101.0 mL, but, as we have already seen, we can ignore the volume and simply work with the amounts of acid and base present.

Calculate for an ammonia buffer solution with a pH of 10.2. pK a (NH 4 + ) = 9.24. Write the expression for K a : NH 4 + + H 2 O NH 3 + H 2 O Rearrange it to form the required ratio:

Calculate K a and [H 3 O + ] and hence calculate the ratio.

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