Answer to Question #93857 in General Chemistry for Kylie

Question

(4) What is the molarity of a solution made by dissolving 5.25 g of potassium
hydrogen phthalate in distilled water and making it up to 250 mL with
distilled water in a volumetric flask?

(5) If 16.41 mL of aqueous NaOH is required to neutralise 20 mL of potas-
sium hydrogen phthalate solution described in question 4 above, what is
the molarity of the aqueous sodium hydroxide? Show your calculations
as required below:
(a) Write the reaction equation for the titration.
(b) Calculate the number of moles of KHC8
H4O4
used in the titration.
(c) Using the balance reaction equation, find the number of moles
of NaOH that reacted.
(d) Calculate the molarity of the NaOH solution.

Answer 1

(4)

$$c = n(KHP)/V_{solution}$$

$$n(KHP) = m/M$$

If the molecular formula is potassium hydrogen phthalate $C_4H_5KO_4$ , then:

$$M(KHP) = 204.222 g/mol$$

Eventually:

$$n(KHP) = {5.250g \over 204.222g/mol} = 0.026 mol$$

$$c = {0.026mol \over 0.250L} = 0.104 mol/L$$

(5)

a)

$$KHP + NaOH = NaKP + H_2O$$

b), c), d)

$$n(KHP) = n(NaOH)$$

$$n(NaOH) = 0.026 mol$$

$$c(NaOH) = n(NaOH)/V_{solution}$$

$$c(NaOH) = {0.026 mol\over 0.020L} = 1.3 mol/L$$

Or:

$$c(KHP)V(KHP) = c(NaOH)V(NaOH)$$

$$c(NaOH) = {c(KHP)V(KHP)\over V(NaOH)}$$

$$c(NaOH) = {0.104 \cdotp 250 \over20} = 1.3 mol/L$$