Answer to Question #325200 in General Chemistry for Mut

Question

2. A 0.2719 g sample containing

a reacted with 20.00 mL of 0.2254 M HCI.

Given that HCI was excess. The excess HCI required exactly 20.00 mL of 0.1041

M NaOH to reach the end-point using phenolphthalein indicator.

Determine percentage purity of

3() +2

( ) -

2( )+2 20 (1)

The titration reaction is

a in the sample. The reaction involved is

( ) + NaOH(aq)

CI( ) + 2 (1)

Answer 1

NaOH + HCl = NaCl + H₂O

V(NaOH) = 20.00 mL = 0.02 L;

C(NaOH) = 0.1041 M;

n(NaOH) = C(NaOH) * V(NaOH) = 0.1041 * 0.02 = 0.002082 mol

n(HCl)_{neutralization} = n(NaOH) = 0.002082 mol;

V(HCl)_total = 20.00 mL = 0.02 L;

C(HCl)_total = 0.2254 M;

n(HCl)_total = C(HCl)_total * V(HCl)_total = 0.2254 * 0.02 = 0.004508 mol;

For reaction HCl with sample:

n(HCl)' = n(HCl)_total - n(HCl)_{neutralization} = 0.004508 - 0.002082 = 0.002426 mol.