A solution is made by mixing 5.00 times 10^2 mL of 0.167 M NaOH with 5.00 times

Question

A solution is made by mixing 5.00 times 10^2 mL of 0.167 M NaOH with 5.00 times 10^2 mL of 0.100 M CH_3COOH. Calculate the equilibrium concentrations of H^+, CH_3COOH, CH_3COO^-, OH^-, and Na^+. Calculate the pH at the equivalence point for the following titration: 0.20 M HCI versus 0.20 M methyl-amine (CH_3NH_2). Calculate the pH at the equivalence point for the following titration: 0.10 M HCOOH versus 0.10 M NaOH. A 25.0-mL solution of 0.100 A/CH, COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: 0.0 mL, 5.0 mL, 10.0 mL, 12.5 mL. 15.0 mL. A 10.0-mL solution of 0.300 M NH_3 is titrated with a 0 100 M HCI solution. Calculate the pH after the following additions of the HCI solution: 0.0 mL, 10.0 mL, 20.0 mL, 30.0 mL, 40.0 mL. The diagrams shown here represent solutions at different states in the titration of a weak acid HA with NaOH. Identify the solution that corresponds to (1) the initial stage before the addition of NaOH, (2) halfway to the equivalence point, (3) the equivalence point, (4) beyond the equivalence point. Is the then less than, ore equal to 7 at the

Explanation / Answer

34. at equivalence point of weakacid,strong base titration

   pH = pka of CH3COOH

pka of CH3COOH = 4.74

pH = 4.74

36.

a) before addition of HCl

pH = 14 -1/2(pkb-logC)

pkb of NH3 = 4.74

C = concentration of NH3 = 0.3

    = 14 - 1/2(4.74-log0.3)

   = 11.37

b)

nO of mol of NH3 = 10/1000*0.3 = 0.003 mol

no of mol of HCl = 10/1000*0.1 = 0.001 mol

pH = 14 - (pkb+log(salt/base))

   = 14 - (4.74 + log(0.001/(0.003-0.001)))

= 9.56

c) nO of mol of NH3 = 10/1000*0.3 = 0.003 mol

no of mol of HCl = 20/1000*0.1 = 0.002 mol

pH = 14 - (pkb+log(salt/base))

   = 14 - (4.74 + log(0.002/(0.003-0.002)))

= 8.96

D) nO of mol of NH3 = 10/1000*0.3 = 0.003 mol

no of mol of HCl = 30/1000*0.1 = 0.003 mol

C = concentration of salt = 0.003/40*1000 = 0.075 M

at equivalence point

pH = 7 -1/2(pkb+logC)

    = 7-1/2(4.74+log0.075)

   = 5.2

e) nO of mol of NH3 = 10/1000*0.3 = 0.003 mol

no of mol of HCl = 40/1000*0.1 = 0.004 mol

C = concentration of excess HCl = (0.004-0.003)/50*1000 = 0.02 M

pH = -log(H+)

= -log0.02

= 1.7

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