Question

A student titrated a weak acid with 0.0994 M NaOH solution. It took the student 34.65mL of NaOH to reach the equivalence point. How many grams of weak acid did the student use for titration, if molar mass of the weak acid was 105.125 g/mol

Answer 1

0.362 g of weak acid was used for titration

Step-by-step explanation:

Let's assume the weak acid (HA) is monobasic.

Balanced equation:
`HA+NaOH\rightarrow NaA+H_(2)O`

So, 1 mol of NaOH neutralizes 1 mol of HA

Number of moles of NaOH in 34.65 mL of 0.0994 M of NaOH =
`(0.0994* 34.65)/(1000)moles=0.00344moles`

So, 0.00344 moles of NaOH neutralizes 0.00344 moles of HA

So, the student used 0.00344 moles of acid for titration.

So, mass of acid was used =
`(105.125* 0.00344)g=0.362g`