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.


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](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
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
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Answer:
16mL
Explanation:
Using the following formula;
CaVa = CbVb
Where;
Where
Ca = concentration/molarity of acid (M)
Va = volume of acid (mL)
Cb = concentration/molarity of base (M)
Vb = volume of base (mL)
According to the information provided in this question;
Ca (HCl) = 2M
Cb (NaOH) = 5M
Va (HCl) = 40mL
Vb (NaOH) = ?
Using CaVa = CbVb
Vb = CaVa/Cb
Vb = 2 × 40/5
Vb = 80/5
Vb = 16mL
Answer:

Explanation:
Hello,
In this case, for the given reaction:

We find a 1:2 molar ratio between the acid and the base respectively, for that reason, at the equivalence point we find:

That in terms of concentrations and volumes we can compute the concentration of the acid solution:

Best regards.