Please show work Need help A.) A 100.0 mL sample of 0.10 M NH 3 is titrated with
ID: 965283 • Letter: P
Question
Please show work Need help
A.) A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 50.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.
A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 50.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.
D.)What is the pH at the equivalence point of a weak base-strong acid titration if 20.00 mL of NaOCl requires 28.30 mL of 0.10 M HCl? K a = 3.0 × 10-8 for HOCl.
What is the pH at the equivalence point of a weak base-strong acid titration if 20.00 mL of NaOCl requires 28.30 mL of 0.10 M HCl? K a = 3.0 × 10-8 for HOCl.
7.05 9.26 4.74 10.34 7.78Explanation / Answer
A ) Total volume of solution = 100 ml Ammonia + 50 ml Nitric acid = 150 mL
In this solution,50 mL of NH4OH are formed
1) No. of moles of NH3 in 150 ml of solution = 0.10 x 100 / 1000 = 0.01 moles
2) No of moles of NH4OH in 150 mL of solution = 0.10 x 50 / 1000 =0.005 moles
Now,
pOH = -log Kb + log[ salt]/[acid]
= -log 1.8 ×10-5 x log 0.005 / 0.01
= 4.744 - 0.3010 = 4.443
b)Total volume of solution = 40 ml acetic acid = 10 ml NaOH = 50 mL
In this solution,10 mL of CH3COONa are formed
1) No. of moles of CH3COOH in 50 ml of solution = 0.10 x 40/ 1000 = 0.004 moles
2) No of moles of CH3COONa in 50 mL of solution = 0.10 x 10 / 1000 =0.001 moles
Now,
pH = -log Ka + log[ salt]/[acid]
= -log 1.8 ×10-5 x log 0.001 / 0.004
= 4.744 - 0.6020 = 4.14