Question

A student titrates a 22.0mL solution of 0.200M HCN with 0.200M NaOH. The acid dissociation constant of HCN is 6.2 x 10^(-10)

1. Determine the pH of the acetic acid solution before any NaOH is added.
2. Determine the pH of the solution when 6.00mL of 0.100M NaOH has been added
3. Determine the pH of the solution when17.0mL of 0.100M NaOH has been added.
4. Determine the pH of the solution at the equivalence point
5. Determine the pH of the solution after an extra 2.0mL of NaOH is added past the equivalence point.
6. Use the volume of NaOH and pH of the solution to sketch a titration graph for the reaction between NaOH and HCN

  

Answer 1

0 4см Weak alid base salt waten kon ette HCN + Naon - NaCN 7110 Katow = 6.2. xjö ' o pke -logka--leq6zil 1 pka. 9. 2 Na OH Molarity (M, ) =0.2000 M2 = 0.200 Volume (V) : 22.0me V2 ? At vazone for weak acids [ht] = Skal = 58.2x100x 0-200 = lollx105 pH = -log€4+3 = -loglixios - 4.95 ② No = 6ooome 4.95 : D [Acid] = [HCNJ ett = M, N, -H₂ V 2 Vitva = 0.200*22-0-200x6.0 - 22+6 - 0:1143M

Esalt] = [Nacwformed = M2 V2 U+U2 = 0.200 x6.00 6600122 -0.04286 ...pH = pka + log esalt I talid] -9,2+ log 0.04286 - 9.2 -0.426 O 07143 -7 pH = 8:7767 3 2 = 170me [Hon] = 0.2008 22-0:200817 X470-02584 2277 Ewaew] = 0.2.00817 = 0.0872 19.73 pH = 9.2+log 02564 - 9 At Equivalence point o=22me 22+1 0.0872 - present.. [NacN] = 0 200x22 Only Nain is 22722 TUDM for weak alid - strong base salt pH = of [ pkw + pka +log esalts] =4[14.00 +962 +10g0.1] (PAED 6 V2 =22+2=24ml .

© V2 = 24 me i [naoh von = Mq U₂ - MOVI - VAU2 [NaCH] = 0.288x24 -0.288x22 24+22 (NaOH) = 8,695x1 mg [NaOH J = COH ] = 8:695x 103 m pot = 8.06 HipH -14.00 -POH = 14,00-2.06 = 11.94 Using these values we can sketch a titration graph vol. Naoth