Question

Acid Ka H2PO4- 6.29 x 10-8 NH4+ 5.68 x 10 -10 HCl >> 1 ​ Calculate...

Acid Ka

H2PO4- 6.29 x 10-8

NH4+ 5.68 x 10 -10

HCl >> 1 ​

Calculate the pH of a 0.150 M solution of Na2HPO4.

Calculate the pH of a 0.150 M solution of NH4NO3.

Calculate the pH of a 0.150 M solution of KCl.

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Homework Answers

Answer #1

1) Na2HPO4 is salt of weak acid H2PO4- and strong base NaOH

for such salts

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

pKa = - log Ka = - log [6.29 x 10-8] = 7.20

pH = 1/2 [14 + 7.20 + log 0.150]

pH = 10.19

2) NH4NO3 is salt of weak base and strong acid

pKa of NH4+ = - log [5.68 x 10-10] = 9.24

pKb = 14 - 9.24 = 4.76

pOH = 1/2 [pKw + pKb + log C]

pOH = 1/2 [14 + 4.76 + log 0.150]

pOH = 8.97

pH = 14 - 8.97

pH = 5.03

3) KCl is salt of strong acid and strong base

such salts pH always 7.0

pH = 7.0

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