Question

6. (2 Points) Suppose 50.0 mL of 0.350 M lithium hydroxide is mixed with 30.0 mL of 0.250 M perchloric acid. What is the pH of the resulting solution, assume the reaction goes to completion.

Answer 1

Answer:  13.1

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in ml)}}$     .....(1)

Molarity of LiOH solution = 0.350 M

Volume of solution = 50.0 mL

Putting values in equation 1, we get:

$0.350M=\frac{\text{Moles of LiOH}\times 1000}{50.0L}\\\\\text{Moles of LiOH}=\frac{0.350mol/L\times 50.0}{1000}=0.0175mol$

Moles of $OH^-$ ion = 0.0175 moles

$0.250M=\frac{\text{Moles of}HCLO_4\times 1000}{30.0L}\\\\\text{Moles of} HClO_4=\frac{0.250mol/L\times 25.0}{1000}=0.00625mol$

Moles of $H^+$ ion = 0.00625 moles

The chemical equation for the reaction of LiOH with $HClO_4$ follows:

$HClO_4+LiOH\rightarrow LiClO_4+H_2O$

For neutralization:

1 mole of $H^+$ ion will react with 1 mole of $OH^-$ ion

As, 0.00625 moles of $H^+$ ion react with=$\frac{1}{1}\times 0.00625=0.00625$ moles of $OH^-$ ion

Moles of $OH^-$ left = (0.0175-0.00625) = 0.01125 moles

Concentration of $OH^-=\frac{moles}{\text {total volume in L}}=\frac{0.01125}{0.0800L}=0.141M$

$pOH= -log[OH^-]$

$pOH= -log[0.141]=0.851$

pH +pOH = 14

pH = 14- pOH = 14 -0.851 = 13.1

Thus pH of the resulting solution is 13.1