Question

consider the reaction of 75.0 ml of 0.350 m c₅h₅n (kb = 1.7×10⁻⁹) with 100.0 ml of 0.425 m hcl.
with 0.0162 moles of h⁺ in excess in a volume of 175.0 ml, what would be the ph of this solution after the reaction?

Explanation:

Step1: Calculate the concentration of \(H^+\)

The formula for molarity \(M=\frac{n}{V}\), where \(n\) is the number of moles and \(V\) is the volume in liters.
Given \(n = 0.0162\) moles and \(V=175.0\space mL=0.1750\space L\)
\(M=\frac{0.0162}{0.1750}\)

Step2: Calculate the \(pH\)

The formula for \(pH\) is \(pH =-\log[H^+]\)
Substitute the value of \([H^+]\) from step 1 into the formula.

Answer:

First, calculate the concentration of \(H^+\):

$$
LATEXBLOCK0
$$

Then, calculate the \(pH\):

$$
LATEXBLOCK1
$$

So the \(pH\) of the solution is \(1.03\)