Calculate the pOH of a solution that is 0.448 M NH 2 NH 2 and 0.187 M NH 2 NH 3
ID: 495214 • Letter: C
Question
Calculate the pOH of a solution that is 0.448 M NH2NH2and 0.187 M NH2NH3Cl. Kbof NH2NH2is 1.7 x 10-6?
Calculate the pH when 0.89 g of NaF is added to 44 mL of 0.50 M HF. Ignore any changes in volume. The Ka value for HF is 3.5 x 10-4?
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Explanation / Answer
1)
pKb = -log Kb = -log (1.7 x 10-6)
= 5.77
pOH = pKb + log [NH2NH3Cl /NH2NH2 ]
= 5.77 + log (0.187 / 0.448)
pOH = 5.39
2)
Calculate the pH when 0.89 g of NaF is added to 44 mL of 0.50 M HF. Ignore any changes in volume. The Ka value for HF is 3.5 x 10-4?
pKa = 3.46
moles of NaF = 0.89 / 42 = 0.02119
moles of HF = 44 x 0.50 / 1000 = 0.022
pH = pKa + log [NaF / HF]
= 3.46 + log [0.02119 / 0.022]
pH = 3.44
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