Question

A 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution at the equivalence point.

Answer 1

Answer: 7

Step-by-step explanation:

The volume of 0.18 M HClO4 is 100.0 mL.

Calculate the number of moles of 0.18 M HClO4 and 0.27 M LiOH as shown below:

`$$Number of moles $=$ Concentration $*$ Volume$$\begin{aligned}\text { Number of moles of } \mathrm{HClO}_(4) &=100.0 \mathrm{~mL} *\left(\frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}\right) *\left(\frac{0.18 \mathrm{~mol}}{\mathrm{~L}}\right) \\&=0.018 \mathrm{~mol} \\\text { Number of moles of } \mathrm{HClO}_(4) &=66.67 \mathrm{~mL} *\left(\frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}\right) *\left(\frac{0.27 \mathrm{~mol}}{\mathrm{~L}}\right) \\&=0.018 \mathrm{~mol}\end{aligned}$$`

`$$Moles of both acid aand base are equal.\\ $\mathrm{HClO}_(4)$ is a strong acid and $\mathrm{LiOH}$ is a strong base.`

At the equivalence point, the number of moles of the acid completely reacted with all the moles of the added base. Thus, the solution becomes neutral in nature. The pH of a neutral solution is 7. Therefore, the pH of the solution at the equivalence point is 7.