Question

Construct a rough plot of pH versus volume of added base for the titration of 50 mL of 0.070 M HCN with 0.075 M NaOH. Ka = 4.0*10-10 for HCN

(a) What is the pH before any NaOH is added?

(b) What volume of base, in milliliters, is required to reach the equivalence point?

(c) What is the pH at the equivalence point?

(d) What indicator would be most suitable to determine the endpoint?

(1).bromcresol purple

(2).bromocresol green     

(3).Bromthymol blue

(4).Alizarin yellow GG

Answer 1

Construct a onh plot of pH versus volume of added base for the titration of 50 mL of 0.070 M HCN with 0.075 M NaOIH. Ka -4.0 10 10 for HCN Molarity- mumber of moles volume in L. Volume il L = Number ofmoles molarity -0.0035 Moles 0.075 M NaOH -0.0466 L - 46.6 ml (c) What is the pH at the equi valence poit? Moles of acid - molarity * volume in L -0.0708 0.050 - 0.0035 Moles HCN Moles NaOH needed 0.0035 Duc to small value of Ka we can writc 0.070-X 4.0*10-10-х"X/ 0.070 4.0 10-10-X 2/0.070 0.070 10.98 ecutierce point 25 volume cfakal added fam X= 5.3*10^-6 pH =-log[H+] =-log 5.3*10-4 -5.28 (a) What is the pH before any NaOH is added? Volume NaOH = = 0.0035 Moles 10.075 M NaOH -0.0466 L -46.6 Il Here Ka 4.010-10 tor HCN (b) What volue of base, in milliliters. is required to reach the equivalence poil? Ka-[H+][CN-HN] Assume tha H[CN--X Ka -[H+][CN-HCN] Then 4.0#10-10-X*X/ 0.070-X HCN NaOH NaCN H20 total volhme 50.0 m- 46.6 96.6-0.0966 T conccntration CN-0.0035 0.0966 L 0.036 M Kb-Kw/ Ka-1.0#10/-14 /4,0+ 10-10-25% 10/-5 Constict a rough plot of pH versus volume of added base for the titration of 50 mL of 0.070 M IHCN with 0.075 M NaOIH. Ka -4.0 10 10 for HCN Kb-x2 0.036 -x 2.5 * 10^-5 = x^2/ 0.03 6-X 2.5 10A-Sx2/0.036 x OH-1 0.000948 M pOH 3.02 pH 14- pOH-10.98 (d) What indicator would be most suitable to detcrminc the cndpoint? (1).bromcresol purplc (2).bromocresol green Moles of acid = molarity * volume in L -0.070* 0.050 -0.0035 Moles HCN According to reaction moles of base 0.0035 Moles