Chemistry help with

Question

chemist titrates a 25.00 mL sample of 0.1191 M benzoic acid (C6H5COOH) against a 0.1000 M solution of NaOH. The overall reaction is shown by the equation below: What volume of NaOH solution (in mL) is required to reach the equivalence point?

Which of the following quantities will be needed to calculate the pH at the equivalence point of the reaction?

Choose one or more: total solution volume moles C6H5COO– moles C6H5COOH [C6H5COOH]equivalence Kb for C6H5COO–

Using your answers to Parts 1 through 3, calculate the pH at the equivalence point of the reaction. Ka for C6H5COOH is equal to 6.28 × 10–5 and Kb for C6H5COO– is equal to 1.59 × 10–10. Give your answer to two places

Answer 1

C6H5COOH+ NaOH=C6H5COONa+H2O

First, let's find out, how many gram of benzoic acid is in 1 liter of 0,1191M solution.

1 mol of Benzoic acid is 122 gram. But we have 0.1191 mol or 14,53 gram.

That amount is in 1 liter, but we have 25 ml. So, 14,53 gram x 0,025 l=0.3632gram benzoic acid

Now, we can find how many gram NaOH . (0.363x40):122=0,119 gram

0, 1 M NaOH means 4gram in 1L, but we have 0,119 gram NaOH

So, solution of NaOH will be 0.03 L or 300mL