How many mL (to the nearest mL) of 0.188-M KF solution should be added to 530. mL of 0.240-M HF to prepare a pH = 2.40 solution?
? mL KF
As you aware, that pH = 2.40 means [H3O+] = 10^-2.40. Now:
HF + H2O ⇌ H3O+ + F-, pKa = 3.15 (in water) <<The pKa value
is from Wikipedia>>
Ka = 10^-3.15 = [H3O+]*[F-]/[HF] = (10^-2.40)*[F-]/[HF]
Hence: [F-]/[HF] = (10^-3.15)/(10^-2.40) = 0.177
Assume X ml of 0.188 M KF solution should be added. Once it is
mixed with 530. mL of 0.240 M HF,
[F-] = 0.188*X/(X+530), and
[HF] = 0.240*530/(X+530). Hence:
[F-]/[HF] = 0.177 = 0.188*X/(0.240*530)
X = 119.75ml = .119L
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