36.
According Henderson - Hasselbaltch's equation of acid buffers,
pH = pKa + Log[Salt] / [Acid]
2.75 = 3.75 + Log[HCOO-]/[HCOOH]
2.75 - 3.75 = Log[HCOO-]/[HCOOH]
- 1.00 = Log[HCOO-] / [HCOOH]
[HCOO-] / [HCOOH] = 1.00 x 10^-1
37.
Initial moles of acid = molarity x volume in L = 0.181 x 1.42 = 0.257 mol
Initial moles of salt = 0.310 x 1.42 = 0.440 mol
Addition of a base, NaOH increases the concentration of salt and decreases the concentration of acid.
SO,
Final moles of acid = 0.257 - 0.069 = 0.188 mol
Final moles of salt = 0.440 + 0.069 = 0.509 mol
Ka = 1.52 x 10^-5
pKa = - LogKa = - Log(1.52 x 10^-5) = 4.82
Formula,
pH = pKa + Log[salt]/[acid[
pH = 4.82 + Log(0.509 / 0.188)
pH = 4.82 + 0.432
pH = 5.25
38.
Let us consider the volume of acid = x mL
Then,
Volume of salt = 100 - x mL
pH = pKa + Log[salt]/[acid[
4.00 = 4.20 + Log(0.140 (100 - x) / (0.100 x)
Log0.140(100-x) / (0.100 x ) = - 0.20
0.140 ( 100 - x ) / 0.100 x = 10^-0.20
14.0 - 0.140 x = 0.631 * 0.100 x
14.0 = 0.0631 x + 0.140 x
14.0 = 0.203 x
x = 14.0 / 0.203
x = 69.0 mL
Therefore,
Volume of acid solution = 69.0 mL
Volume of salt solution = 100 - 69.0 = 31.0 mL
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