Question

Question 2

1. Boric acid has a K_a of 5.8 × 10^-10. What is the pKa? Assume boric acid is H₃BO₃ and is monoprotic.

Question 3

1. A 0.25 M solution of boric acid is mixed with a 0.75 M solution of NaOH. What is the pH of the solution if 7.5 mL of boric acid is mixed with 5 mL of NaOH?

Question 4

1. What volume of boric acid solution (in mL) is required to fully neutralize the sodium hydroxide solution?

Answer 1

2)

Given:

Ka = 5.8*10^-10

use:

pKa = -log Ka

= -log (5.8*10^-10)

= 9.24

Answer: 9.24

3)

Given:

M(H3BO3) = 0.25 M

V(H3BO3) = 7.5 mL

M(NaOH) = 0.75 M

V(NaOH) = 5 mL

mol(H3BO3) = M(H3BO3) * V(H3BO3)

mol(H3BO3) = 0.25 M * 7.5 mL = 1.875 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.75 M * 5 mL = 3.75 mmol

We have:

mol(H3BO3) = 1.875 mmol

mol(NaOH) = 3.75 mmol

1.875 mmol of both will react

excess NaOH remaining = 1.875 mmol

Volume of Solution = 7.5 + 5 = 12.5 mL

[OH-] = 1.875 mmol/12.5 mL = 0.15 M

use:

pOH = -log [OH-]

= -log (0.15)

= 0.8239

use:

PH = 14 - pOH

= 14 - 0.8239

= 13.1761

Answer: 13.18

4)

Balanced chemical equation is:

NaOH + H3BO3 ---> H2BO3- + H2O

Here:

M(NaOH)=0.75 M

M(H3BO3)=0.25 M

V(NaOH)=5.0 mL

According to balanced reaction:

1*number of mol of NaOH =1*number of mol of H3BO3

1*M(NaOH)*V(NaOH) =1*M(H3BO3)*V(H3BO3)

1*0.75 M *5.0 mL = 1*0.25M *V(H3BO3)

V(H3BO3) = 15.0 mL

Answer: 15.0 mL