Question

In: Chemistry

1) A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate...

1) A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 23.0 mL of HNO3.

2)A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 21.0 mL of NaOH.

Solutions

Expert Solution

Part.1

Initial moles of NH3 = 0.075 L x 0.200 M = 0.015 mol

Moles of HNO3 added = 0.023 L  x 0.500 M = 0.0115 mol

NH3 + HNO3 ----------> NH4+ + NO3-

Moles of NH3 left = 0.015 - 0.0115 = 0.0035 mol

Moles of NH4+ = 0.0115 mol

Ka(NH4+) = Kw/Kb(NH3)

= 10-14/1.8 x 10-5 = 5.556 x 10-10

Henderson-Hasselbalch equation:

pH = pKa + log([NH3]/[NH4+])

= - log Ka + log 0.0035/0.0115

= -log(5.556 x 10-10) + log 0.0035/0.0115

= 9.26 + log 0.3043

= 9.26 - 0.5167

pH = 8.74

Part.2

Initial moles of CH3COOH = 0.052 L x 0.35 M = 0.0182 mol

Moles of NaOH added = 0.021 L x 0.400 M = 0.0084 mol

CH3COOH + NaOH ---------> CH3COO- + Na+

Moles of CH3COOH left = 0.0182 - 0.0084 = 0.0098 mol

Moles of CH3COO- = 0.0084 mol

Henderson-Hasselbalch equation:

pH = pKa + log([CH3COO-]/[CH3COOH])

= -log Ka + log([CH3COO-]/[CH3COOH])

= -log(1.8 x 10-5) + log(0.0084/0.0098)

= 4.74 + log 0.8571

= 4.74 - 0.06697

pH = 4.67


Related Solutions

Part A A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3....
Part A A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 25.0 mL of HNO3. Part B A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 21.0 mL of NaOH.
Part C A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3....
Part C A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 23.0 mL of HNO3. Part D A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 MNaOH. Calculate the pH after the addition of 17.0 mL of NaOH. please show work
Part A: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5 ) is titrated with 0.500 M...
Part A: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5 ) is titrated with 0.500 M HNO3 . Calculate the pH after the addition of 28.0 mL of HNO3 . Part B: A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5 ) is titrated with 0.40 M NaOH . Calculate the pH after the addition of 33.0 mL of NaOH . Please Explain Steps
Part C A 75.0-mL volume of 0.200 mol L−1 NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 mol...
Part C A 75.0-mL volume of 0.200 mol L−1 NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 mol L−1 HNO3. Calculate the pH after the addition of 28.0 mL of HNO3. Express your answer numerically. Part D A 52.0 mL volume of 0.350 mol L−1 CH3COOH (Ka=1.8×10−5Ka=1.8×10−5) is titrated with 0.400 mol L−1 NaOH. Calculate the pH after the addition of 15.0 mL of NaOH. Express your answer numerically.
A. A 75.0-mLmL volume of 0.200 MM NH3NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 MM HNO3HNO3. Calculate...
A. A 75.0-mLmL volume of 0.200 MM NH3NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 MM HNO3HNO3. Calculate the pHpH after the addition of 11.0 mLmL of HNO3HNO3. B. A 50.0-mLmL volume of 0.15 MM HBrHBr is titrated with 0.25 MM KOHKOH. Calculate the pHpH after the addition of 15.0 mLmL of KOHKOH. C. A 52.0-mLmL volume of 0.35 MM CH3COOHCH3COOH (Ka=1.8×10−5Ka=1.8×10−5) is titrated with 0.40 MM NaOHNaOH. Calculate the pHpH after the addition of 17.0 mLmL of NaOHNaOH.
A 35.0 mL sample of 0.150 M ammonia (NH3, Kb=1.8×10−5) is titrated with 0.150 M HNO3....
A 35.0 mL sample of 0.150 M ammonia (NH3, Kb=1.8×10−5) is titrated with 0.150 M HNO3. Calculate the pH after the addition of each of the following volumes of acid. Express your answer using two decimal places. A. 0.0 mL B. 17.5 mL C. 35.0 mL D. 80.0 mL
1. A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Kb,...
1. A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Kb, NH3 = 1.8 × 10−5. Determine the pH of the solution at each of the following points in the titration: (a) before addition of any HNO3 (b) after the addition of 50.0 mL HNO3 (c) after the addition of 75.0 mL HNO3 (d) at the equivalence point (e) after the addition of 150.0 mL HNO3 2. (12 points) Consider the titration of 37.0 mL...
Take 12.5 mL of 0.200 moles of aqueous Ammonia (NH3), whose Kb = 1.8 x 10-5,...
Take 12.5 mL of 0.200 moles of aqueous Ammonia (NH3), whose Kb = 1.8 x 10-5, and Titrate with 0.100 mol HCl, calculate PH at 5 mL HCl, 10 mL , 15 mL, 20 mL, 24.5mL, 25mL, 25.5mL, 30mL 40.0 mL, abd 50 mL HCl. Graph the solution with hoizontal axis being the PH up to 14 and The y-axis being mL HCl.
Consider the titration of 40.00 mL of 0.200 M NH3 with 0.500 M HCl. Calculate pH...
Consider the titration of 40.00 mL of 0.200 M NH3 with 0.500 M HCl. Calculate pH a) initially   b) after addition of 5.00 mL of 0.500 M HCl c) at the half-neutralization point (halfway to equilvalence point) d) after the addition of 10.00 mL of 0.500 M HCl e) at the equivalence point
Given that Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5 , calculate Kb...
Given that Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5 , calculate Kb for CN− and Ka for NH4+ . Enter the Kb value for CN− followed by the Ka value for NH4+ , separated by a comma, using two significant figures.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT