1. A solution contains 0.285 M sodium acetate and 8.74×10-2 M acetic acid. The p

Question

1. A solution contains 0.285 M sodium acetate and 8.74×10-2 M acetic acid. The pH of this solution is

2.A solution contains 0.245 M potassium hypochlorite and 0.388 M hypochlorous acid. The pH of this solution is

3.The compound ethylamine is a weak base like ammonia. A solution contains 0.392 M C2H5NH3+ and 0.381 M ethylamine, C2H5NH2. The pH of this solution is

4. A solution contains 6.06×10-2 M ammonium chloride and 0.447 M ammonia. The pH of this solution is

5.A buffer solution is 0.385 M in HNO2 and 0.259 M in KNO2. If Ka for HNO2 is 4.5×10-4, what is the pH of this buffer solution?

6.A buffer solution is 0.305 M in H2C2O4 and 0.223 M in NaHC2O4. If Ka for H2C2O4 is 5.9×10-2, what is the pH of this buffer solution?

Explanation / Answer

Using Henderson Hasselbalch equation

pH = pKa + log { [Salt] / [Acid] }

pOH = pKb + log { [Salt] / [Base] }

1.

pH = pKa + log { [Salt] / [Acid] }

pka of acetic acid = 4.75

[Salt] = [sodium acetate] = 0.285 M

[Acid] = [acetic acid] = 8.74 x 10-2 M = 0.0874 M

So, pH = 4.75 + log ( 0.285 M / 0.0874 M )

= 4.75 + 0.51

= 5.26

2.

pH = pKa + log { [Salt] / [Acid] }

pka of hypochlorous acid = 7.40

[Salt] = [potassium hypochlorite] = 0.245 M

[Acid] = [hypochlorous acid] = 0.388 M

So, pH = 7.40 + log ( 0.245 M / 0.388 M )

= 7.40 - 0.20

= 7.20

3.

A solution contains 0.392 M C2H5NH3+ and 0.381 M ethylamine, C2H5NH2. The pH of this solution is

pOH = pKb + log { [Salt] / [Base] }

pkb of ethylamine, C2H5NH2 = 3.19

[Salt] = [C2H5NH3+] = 0.392 M

[Base] = [C2H5NH2] = 0.381 M

So, pOH = 3.19 + log ( 0.392 M / 0.381 M )

= 3.19 + 0.01

= 3.20

pH = 14 - pOH = 14 - 3.20 = 10.8

4.

pOH = pKb + log { [Salt] / [Base] }

pkb of ammonia = 4.75

[Salt] = [ammonium chloride] = 6.06 x 10-2 M = 0.0606 M

[Base] = [ammonia] = 0.447 M

So, pOH = 4.75 + log ( 0.0606 M / 0.447 M )

= 4.75 - 0.87

= 3.88

pH = 14 - pOH = 14 - 3.88 = 10.12

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