Question 2
Boric acid has a Ka of 5.8 × 10-10. What is the pKa? Assume boric acid is H3BO3 and is monoprotic.
Question 3
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
What volume of boric acid solution (in mL) is required to fully neutralize the sodium hydroxide solution?
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
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