Question

Question 2.


A. A student titrated 10ml of 0.200M NaOH against an unknown concentration of hydrocholric acid (HCl) using phenolpthalein as an indicator. 10ml of acid was required to obtain the first permanent colour change.


i. Write the net ionic equation for this titration


ii. What was the colour change at the end point


iii. Calculate in moles/hydroxide ions originally present in the titration


iv. Calculate the unknown concentration of the hydrochloric acid


B. A 0.200M solution of methanoic acid (HCOOH) has a Ka value of 1.82 x 10⁻4


i. Calculate the equilibrium concentration of this solution


ii. Calculate the pH of the solution

Answer 1

A

i) H⁺ + OH^- = H₂O

ii) The indicator changed the colour from pink to transparent

iii) n(OH-) = n(NaOH) = c(NaOH)*V(NaOH) = 0.2 mol/L*0.01 L = 0.002 mol

iv) n(HCl) = n(NaOH) = 0.002 mol; c(HCl) = n(HCl)/V(HCl) = 0.002 mol/0.01 L = 0.2 mol/L

B

i) HCOOH <-> H⁺ +HCOO^-

K_a = c(H⁺)*c(HCOO^-)/c(HCOOH)

Concentration of H⁺ ions equals x, concentration of formate anions also equals x, concentration of methanoic acid equals (0.2-x)

K_a = x*x/(0.2-x) = 1.82*10^-4

x = 1.18*10^-2

c(HCOOH) = 0.2-x = 0.188 mol/L

ii) pH = -lgC(H⁺)

c(H⁺) = x = 1.18*10^-2

pH = 1.93