Question

determine the pH at each stage: (a) before adding 0.50 M of NaOH to 25.0 mL of 0.45 M HCl (b) after adding 15.00 mL of NaOH to 25.00 mL of 0.45 M of HCl. (c) at the equivalent point (D) after adding 35.00 mL 0.50 M of NaOH to 25.00 mL of 0.45 M HCl

Answer 1

(a)

before adding 0.50 M of NaOH to 25.0 mL of 0.45 M HCl

Molarity of HCl = moles of HCl / volume on Liter

Moles = 0.45 moles / 1.00 L * 0.025 L=

1.125*10^-5 moles

M = 1.125*10^-5 moles/ Volume in L

= 4.5*10^-4 M

HCl = H+ +Cl-

4.5*10^-4 M = HCl = H+

pH = -log [H+]= -log 4.5*10^-4 M

= 3.35

(b) after adding 15.00 mL of NaOH to 25.00 mL of 0.45 M of HCl

HCl + NaOH --> H2O + NaCl

See which one has more moles:

#mol HCl = 0.45M * 0.025L = 0.01125 mol HCl
#mol NaOH = 0.50 * 0.015L = 0.0075 mol NaOH

HCl clearly has the more moles.
Thus, the excess number of moles of HCl is
0.01125-0.0075 = 0.00375 mol

#M HCl = 0.00375 mol/ 0.025+0.015 L = = 0.09375 M HCl
(note: 0.040 L used since total volume of both solutions is 40mL = 0.04L)

HCl = H+ +Cl-

0.09375 M = HCl = H+

pH = -log [H+]= -log 0.09375 M

= 1.028

(c) at the equivalent point

The reaction of a strong base such as NaOH with a strong acid such as HCl gives a salt which is neutral:
NaOH + HCl → NaCl + H2O
NaCl is neutral. The equivalence point is at neutral pH = 7.00.

(D) after adding 35.00 mL 0.50 M of NaOH to 25.00 mL of 0.45 M HCl

HCl + NaOH --> H2O + NaCl

See which one has more moles:

#mol HCl = 0.45M * 0.025L = 0.01125 mol HCl
#mol NaOH = 0.50 * 0.035 L = 0.0175 mol NaOH

NaOH clearly has the more moles.
Thus, the excess number of moles of NaOH is
0.0175 -0.01125 = 0.00675 mol

#M NaOH = 0.00675 mol/ 0.025+0.035 L = = 0.104M NaOH
(note: 0.060 L used since total volume of both solutions is 60mL = 0.06L)

NaOH = Na+   OH-

0.104M = NaOH =    OH-

pOH = -log [OH-]= -log 0.104M

= 0.98

pH = 14-pOH

pH = 14-0.98

=13.02