b. A 1.00 L buffer solution is 0.150 M in HC7H5O2 and 0.250 M in LiC7H502....

Question

Question

b. A 1.00 L buffer solution is 0.150 M in HC7H5O2 and 0.250 M in LiC7H502. Calc 52 and 0.250 M in LiC7H502. Calculate the pH

Answer 1

Answer #1

Moles of LiC₇H₅O₂ = molarity × volume = 0.150 × 1.00 = 0.1500

Moles of HC₇H₅O₂ = 0.250×1.00 = 0.250

Moles of HCl = 0.10

Reaction when HCl is added

HCl + ^-C₇H₅O₂ $\to$ HC₇H₅O₂ + Cl^-

BCA table is

Now,

pKa = - log Ka = - log(6.5×10^-5) = 4.18

pH using Henderson-Hasselbalch equation is

pH = pKa + log [^-C₇H₅O₂]/[C₇H₅O₂]

So, pH = 4.18 + log (0.05/0.350)

Or, pH = 4.18 - 0.85

Or, pH = 3.33

Answer 2

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