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buffers Q

Ka for methanoic acid is 1.78 × 10−4 mol dm−3 at 25°C.
buffer is prepared containing 2.35 × 10−2 mol of methanoic acid and 1.84 × 10−2 mol of sodium methanoate in 1.00 dm3 of solution.
(i) Calculate the pH of this buffer solution at 25°C.
I did this correct and worked it out as 3.64

however the next part I don't understand how to approach...
A 5.00 cm3 sample of 0.100 mol dm−3 hydrochloric acid is added to the buffer solution in part (c)(i).
Calculate the pH of the buffer solution after this addition.

I did the moles of H+ added as 5x10-4 but that's all I've got so far
Original post by medapplicant2020
Ka for methanoic acid is 1.78 × 10−4 mol dm−3 at 25°C.
buffer is prepared containing 2.35 × 10−2 mol of methanoic acid and 1.84 × 10−2 mol of sodium methanoate in 1.00 dm3 of solution.
(i) Calculate the pH of this buffer solution at 25°C.
I did this correct and worked it out as 3.64

however the next part I don't understand how to approach...
A 5.00 cm3 sample of 0.100 mol dm−3 hydrochloric acid is added to the buffer solution in part (c)(i).
Calculate the pH of the buffer solution after this addition.

I did the moles of H+ added as 5x10-4 but that's all I've got so far


The moles of acid added will interact with the conjugate base ions making more weak acid and removing moles of conjugate base.
Addition of strong acid to the buffer causes the following to happen: H+ + A- \rightarrow HA
Where H- is strong acid, in this case hydrochloric acid; A- is salt, in this case sodium methanoate; and HA is weak acid, in this case methanoic acid

Calculate the moles of hydrochloric acid added to the system, and hence determine how much sodium methanoate reacted and consequently how much methanoic acid was produced (the stoichiometry is 1:1, so the amount of sodium methanoate that reacted = the amount of methanoic acid that was produced)

This will leave you with new concentrations of both species, and since the buffer is of a volume of 1 dm3, the moles of each species is simply equal to their concentrations
You can therefore use the Henderson-Hasselbalch equation to determine how the pH changed due to this addition: pH=pKa+log[HCOO][HCOOH]pH = pKa + log\frac{[HCOO^-]}{[HCOOH]}
(edited 4 years ago)

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