Question

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

Answer 1

Answer: The pH of the resulting solution is 13.1

Explanation:

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

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

Molarity of $LiOH$ solution = 0.350 M

Volume of solution = 50.0 mL

Putting values in equation 1, we get:

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

1 mole of $LiOH$ contains = 1 mol of $OH^-$

Thus $0.0175mol$ of $LiOH$ contain= $\frac{1}{1}\times 0.0175=0.0175$ mol of $OH^-$

$0.250M=\frac{\text{Moles of} HClO_4\times 1000}{30.0ml}\\\\\text{Moles of }HClO_4=\frac{0.250mol/L\times 30.0}{1000}=0.0075mol$

1 mole of $HClO_4$ contains = 1 mol of $H^+$

Thus $0.0075mol$ of $HClO_4$ contain= $\frac{1}{1}\times 0.0075=0.0075$ mol of $H^+$

$HClO_4+NaOH\rightarrow NaClO_4+H_2O$

As 1 mole of $H^+$ combines with 1 mole of $OH^-$

0.0075 moles of $H^+$ combines with = 0.0075 mole of $OH^-$

Moles of $OH^-$ left = (0.0175-0.0075) = 0.01

Moles of $OH^-$ left = $\frac{moles}{Volume}=\frac{0.01}{80.0}\times 1000=0.125M$

pOH =$-log[OH^-]$

pOH =$-log[0.125]=0.90$

$pH+pOH=14$

$pH=14-0.90=13.1$

Thus the ph of the resulting solution is 13.1