Chem 111 help please

Question

If 17.50mL of a NaOH solution is required to titrate a 0.4005g sample of KHP dissolved in 20mL of total solution volume, then what is the molarity of the sodium hydroxide solution?

Thanks in advance!

Answer 1

One problem with using abbreviations, is that they may have more than one meaning. KHP can stand for potassium hydrogen phosphate, or more likely in this case, it stands for potassium hydrogen phthalate (a monoprotic acid). So, assuming KHP is potassium hydrogen phthalate, we have the following reaction:

NaOH + C₈H₅KO₄ ==> H₂O + C₈H₄NaKO₄

molar mass KHP = 204 g/mole

moles KHP present = 0.4005 g x 1 mole/204 g = 0.0001963 moles

moles NaOH needed to neutralize the KHP = 0.0001963 moles (1:1 mole ratio in the balanced equation)

molarity of the NaOH solution = 0.0001963 moles/0.01750 L = 0.11218 M = 0.1122 M (to 4 sig. figs.)

(NOTE: the value of 20 ml used for the KHP solution is not relevant and so is not used in the calculations)

An alternate way of solving this problem, but one that is NOT recommended, would be to use M1V1 = M2V2.

M1 = molarity of KHP = 0.0001963 moles/0.020 L = 0.009815 M

V1 = 20 ml

M2 = ?

V2 = 17.50 ml

M2 = M1V1/V2 = (0.009815)(20)/17.50 = 0.01122 M = 0.01 M (to 1 sig. fig. because 20 ml has only 1 sig. fig.)