1. what is the pH at the equivalence point when 100mL of 0.1 M HNO2 solution (We

Question

1. what is the pH at the equivalence point when 100mL of 0.1 M HNO2 solution (Weak Acid) is titrated with 100 mL of 0.1 M NaOH (Strong Base)?
2. calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
3. what is the molar solubility of AgCl in 0.10 M NH3 at 25oC?
1. what is the pH at the equivalence point when 100mL of 0.1 M HNO2 solution (Weak Acid) is titrated with 100 mL of 0.1 M NaOH (Strong Base)?
2. calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
3. what is the molar solubility of AgCl in 0.10 M NH3 at 25oC?
1. what is the pH at the equivalence point when 100mL of 0.1 M HNO2 solution (Weak Acid) is titrated with 100 mL of 0.1 M NaOH (Strong Base)?
2. calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
3. what is the molar solubility of AgCl in 0.10 M NH3 at 25oC?

Explanation / Answer

1)

HNO2 + NaOH -----------------> NaNO2 + H2O

at equivalence point only salt NaNO2 remains

salt molarity = C = 100 x 0.1 / 100 + 100 = 0.05 M

pKa of HNO2 = 3.25

pH = 7 + 1/2 [pKa + logC]

pH = 7 + 1/2 [3.25 + log 0.05]

pH = 7.97

2)

initial pH :

pOH = pKb + log [NH4Cl / NH3 ]

pOH = 4.74 + log (0.36 x 80 / 0.30 x 80 )

pOH = 4.82

pH + pOH = 14

pH = 9.18

after based addition pH:

pOH = pKb + log [NH4Cl -C / NH3 + C]

pOH = 4.74 + log (0.36 x 80 - 20 x 0.05 / 0.30 x 80 + 20 x 0.05)

pOH = 4.79

pH + pOH = 14

pH = 9.21

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