Question

Calculate the pH of a solution prepared by mixing a) 25.0 mL of 0.512 M NaOH and 34.0 mL of 0.187 M HCl b) 46.0 mL of 0.235 M KOH and 50.0 mL of 0.420 M HC3H5O2 c) 400 mL of 0.250 M NH3 and 250 mL of 0.120 M HCl

Answer 1

1) 25.0 mL of 0.512 M NaOH = 0.0128 moles of NaOH (volume * molarity)

34.0 mL of 0.187 M HCl = 0.00636 moles of HCl    (volume * molarity)

As we can see there is less concentration of HCl, all of it will be neutralized by an equimolar amount of NaOH.

HCl(aq) + NaOH(aq) ---------> H2O(aq) + NaCl(aq)

initial 0.00636 0.1028 0

equilibrium   0.0064     0.00636


0.0128 moles of NaOH - 0.00636 moles of NaOH = 0.0064 excess moles NaOH remaining after reaction.

Divide 0.0064 moles of NaOH by the total volume (0.034 + 0.025)lit to determine the new concentration of the solution formed

0.0064 / 0.0590 = 0.11 M

AS

pOH = -log[OH-] = -log(0.11) = 0.962

pH= 13

2)

46 ml of 0.235 M KOH = 0.046 * 0.235 = 0.01081 Moles of  KOH

50.0 mL of 0.420 M HC3H5O2 = 0.050 * 0.420 = 0.021 moles of HC3H5O2

as we can see that the concentration of HC3H5O2 is in excess

=  0.021 - 0.01081= 0.01019 moles of HC3H5O2

totl volume of new solution after mixing = 46 +50 = 96 ml   

HC2H3O2(aq) + KOH(aq) ------> C3H5O2–(aq) + H2O(l)

Initial concentration   0.021    0.01081 0 0

equillibrium concentration   0.01019   0.01081   

[HC2H3O2] =     0.01019 / 0.096 [ C3H5O2–] =    0.01081 / 0.096 = 0.1126

= 0.10615   

HC3H5O2 --------> H+ + C3H5O2–   

initial conc   0.10615 0   0.1126  

equilibrium conc 0.10615-x +x 0.1126 +x

Ka = [H+ ] [C3H5O2– ] / [  HC3H5O2   ]

1.3 * 10–5 = [x] [0.1126 +x ] /  0.10615-x assuming x very smalll

1.3 * 10–5 = x * [0.1126 ] / [0.10615]

0.1379 * 10–5 = x   [0.1126 ]

1.224 *10-5 = x

As pH = - log [H+] = -log [1.224 *10-5] = 4.9122

3)

moles NH3 = 0.250 mol L * (0.400 L) = 0.100 mol of NH3

moles HCl = 0.120 mol L* (0.250 L) = 0.0300 mol of HCL

NH3(aq) + HCl(aq) --------> NH4Cl(aq)

initial   0.10 0.03

erquilibirium 0.07 0 0.07

[NH3] = 0.070 mol/ 0.650 L = 0.108 M [NH4+] = 0.0300 mol / 0.650 L = 0.0462 M

H2O(l) + NH3(aq) ----------> OH–(aq) + NH4+(aq)

initial 0.108   0.0462

equilibrium 0.108 –x 0.0462+x

Kb= 1.8 x 10–5 = [NH4+][OH–] ? [NH3]

1.8 x 10–5 = [x][0.0462+x] [0.108–x] as x<<<0.0154

4.2 x 10–5 = x = [OH–

] pOH = - log [OH-]= 4.38

as pH= 14- pOH= 14 - 4.38

pH = 9.62