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

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 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* 1000}{50.0ml}\\\\\text{Moles of }LiOH=(0.350mol/L* 50.0)/(1000)=0.0175mol`

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

Thus
`0.0175mol` of
`LiOH` contain=
`(1)/(1)* 0.0175=0.0175` mol of
`OH^-`

`0.250M=\frac{\text{Moles of} HClO_4* 1000}{30.0ml}\\\\\text{Moles of }HClO_4=(0.250mol/L* 30.0)/(1000)=0.0075mol`

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

Thus
`0.0075mol` of
`HClO_4` contain=
`(1)/(1)* 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 =
`(moles)/(Volume)=(0.01)/(80.0)* 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