49.1k views
1 vote
A solution containing 75.0 ml of 0.150 m hcl is titrated with 75.0 ml of 0.300 m naoh. what is the ph of the resulting solution?

User DanHabib
by
7.2k points

1 Answer

5 votes

Final answer:

To calculate the pH after titrating 0.150 M HCl with 0.300 M NaOH, we determine the moles of each, find the excess NaOH, calculate the pOH, and then subtract from 14 to find the pH.

Step-by-step explanation:

To determine the pH of the resulting solution after titration of HCl with NaOH, we need to consider the neutralization reaction that occurs between an acid and a base. In this case, the hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O).

The reaction is as follows: HCl + NaOH → NaCl + H2O.

We first calculate the moles of HCl and NaOH used:
HCl: 75.0 mL × 0.150 M = 0.01125 mol
NaOH: 75.0 mL × 0.300 M = 0.0225 mol

Since NaOH is in excess (0.0225 mol - 0.01125 mol = 0.01125 mol excess NaOH), we then calculate the concentration of OH− ions in the final solution (150.0 mL total volume):
[OH−] = 0.01125 moles / 0.150 L = 0.075 M

We can find the pOH of the solution by taking the negative logarithm (base 10) of the OH− ion concentration: pOH = -log(0.075).

After calculating pOH, we use the relationship pH + pOH = 14 to find the pH. Therefore, pH = 14 - pOH.

User Bob Kerns
by
8.2k points