Question

1) A solution of a weak monoprotic acid of unknown concentration was titrated with 0.23 ​M NaOH. If a 100.-mL sample of the acid solution required exactly 10. mL of the NaOH solution to reach the equivalence point, what was the original concentration of the weak acid?

2) During the titration on problem (​2B​), after 5.0 mL of NaOH addition, the pH = 3.68. What is the K​a​ of the weak acid? please show steps i have an exam tomorrow

Answer 1

1. It is given that it is the titration of weak monoprotic acid with 0.23 M NaOH. The general reaction of a monoprotic acid with NaOH can be written as

MH + NaOH $\rightarrow$ MNa + H₂O

1 mole of monoprotic acid reacts with 1 mole of NaOH for neutralization.

Given that 100 mL acid solution is required to react with 10 mL of 0.23 M NaOH. This means the number of moles of NaOH in 10 mL NaOH solution is equal to number of moles of acid in 100 mL acid solution.

The number of moles of NaOH in 10mL of 0.23M NaOH solution can be calculated from the formula of molarity

Molarity = No. of moles x 1000 / Volume of solution in mL

No. of moles of NaOH = Molarity of NaOH x Volume of NaOH in mL / 1000

No. of moles of NaOH = 0.23 x 10 / 1000 = 0.0023 moles

We know that 1 mole of NaOH reacts with 1 mole of monoprotic acid then

0.0023 moles of NaOH reacts with 0.0023 moles of monoprotic acid

Thus we have the number of moles of acid = 0.0023 moles in 100 mL acid solution

The molarity of the acid solution can be calculated using the same formula used above

Molarity of monoprotic acid = No. of moles of monoprotic acid x 1000 / Volume in mL

Molarity of monoprotic acid = 0.0023 x 1000 / 100 = 0.023 M

Therefore the required concentration of monoprotic acid is 0.023 M