For the two questions below, show detailed calculation setups and results. Pay a

Question

For the two questions below, show detailed calculation setups and results. Pay attention to the correct usage of sig. figs and units. You also will need to look up appropriate constants in Appendix D. Find the pH of the equivalence point(s) and the volume (mL) of 0.125 M HCl needed to reach the point(s) in titrations of (e) 65.5 mL of 0.234 M NH3 (b)21.8 mL of 1.11 MCH,NH2 A 35.00-mL solution of 0.2500 M HF is titrated with a stan- dardized 0.1532 M solution of NaOH at 25°C. (a) What is the pH of the HF solution before titrant is added? (b) How many milliliters of titrant are required to reach the equivalence point? (c) What is the pH at 0.50 mL before the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH at 0.50 mL after the equivalence point?

Explanation / Answer

Q1.

a)

pH of :

a)

mmol of ammonia = MV = 65.5*0.234 = 15.327

Vacid = 15.327 / 0.125 = 122.616

Vtotal = 65.5+122.616= 188.116

[NH4+] = 15.327/188.116 = 0.08147 M

NH4+ = NH3 + H+

Ka = [NH3][H+]/[NH4+]

5.55*10^-10 = x*x/(0.08147 -x)

x = 6.72*10^-6

pH = -log( 6.72*10^-6) 5.17

b)

pKa = 14-3.38 = 10.62

mmol of amine = MV = 21.8*1.11 = 24.198

Vacid = 24.198/ 0.125 = 193.584

Vtotal = 21.8+193.584= 215.384

[CH3NH3+] = 24.198/215.384= 0.1123 M

CH3NH3+ = CH3NH2 + H+

Ka = [CH3NH2 ][H+]/[CH3NH3+]

10^-(10.62) = x*x/(0.1123 -x)

x = 1.64*10^-6

pH = -log( 1.64*10^-6) = 5.78

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