Question

A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titration curve the mixture produced is in, and the pH of the mixture at each volume of added acid. Assume that the volumes of the solutions are additive. Part a): 1) After adding 15.0 mL of the HCl solution, the mixture is the equivalence point on the titration curve. 2) The pH of the solution after adding HCl is

Answer 1

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

$nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol$

$nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol$

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

$[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M$

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

Answer 2

The pH of the solution after adding HCl is mathematically given as

pH= 12.6

Because of excess of NaOH the mixture is before the equivalence point on the titration curve.

What is the position of equivalence point on the titration curve and the pH of the solution after adding HCl ?

Question Parameter(s):

A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH).

Generally, the equation for the  moles of reactants is mathematically given as

 moles of reactants=Molar concentration*Volume

nNaOH=0.25*10*10^{-3}

nNaOH=2.5*10^-3

for

nHCl=0.10 * 15.0* 10^{-3}

nHCl=1.5 * 10^{-3}mol

Because of excess of NaOH the mixture is before the equivalence point on the titration curve.

b)

NaOH  + HCl    ----->       NaCl + H₂O

Therefore

$NaOH=\frac{1.0 * 10^{-3} mol }{25.0*10^{-3} L}\\\\NaOH=0.040M$

Hence

pOH = -log [OH⁻]

pOH= -log 0.040

pOH= 1.4

Giving

pH= 14 - 1.4

pH= 12.6

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