Question

A 5.65 −g sample of a weak acid with Ka=1.3×10−4 was combined with 4.80 mL of 5.90 M NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.25.

Answer 1

what is your question?

do you want molar mass of the acid?

first calculate the no of moles of NaOH = 5.90 M x 0.0048 L = 0.0283 moles

lets un known acid = HA

HA + OH- -----> A- + H2O

no of moles of A- formed = no of moles of NaOH right? = 0.0283 moles

total volume of the solution = 750 mL = 0.75 L

now you have the moles of A- and total vole

Concentration of A- = 0.0283 moles / 0.75L = 0.038M

calculate the pKa value from the given Ka value

pKa = -log(Ka) = -log(1.3×10⁻⁴) = 3.89

resultant solution pH has given 4.25

now use the handerson equation

pH = pKa + log[A-/HA]

4.25 = 3.89 + log[ 0.038/HA]

log[ 0.038/HA] = 4.25-3.89

log[0.038/HA] = 0.36

[0.038/HA] = 10^0.36

[0.038/HA] =2.3

[HA] = 0.0165 moles

we have aMolarity of HA and total volume we can find out the moles of HA

moles HA at equilibirum = 0.0165 M x 0.750 L= 0.0124 moles

intial moles of HA = moles atg equilibrium + moles of A- at equilibrium

initial moles HA = 0.0124 +0.0283 =0.0407 moles of HA

now we have the weight of unknown acid and moles of un known acid

molar mass = weight / moles

= 5.65 g / 0.0407 moles

= 138.82 g/mole