Chemistry, 25.06.2019 09:50 arteria3

A5.36–g sample of nh4cl was added to 25.0 ml of 1.00 m naoh and the resulting solution diluted to 0.100 l. (a) what is the ph of this buffer solution? (b) is the solution acidic or basic? (c) what is the ph of a solution that results when 3.00 ml of 0.034 m hcl is added to the solution?

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Answer from: Lilbre6999

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Explanation:

5.36g NH₄Cl + 25ml(1M NaOH) => (5.36g/53g) NH₄Cl + 0.025L(1M NaOH)

=> 0.101mole NH₄Cl + 0.025mole NaOH

=> (0.101mole/0.025L) NH₄Cl + (0.025mole/0.025L) NaOH

=> 4.045M NH₄Cl + 1.000M NaOH

=> 4.045M NH₄⁺ + 4.045M Clˉ + 0.025M Na⁺ + 0.025M OHˉ

=> 0.025M NH₄OH + (4.045 – 0.025)M NH₄⁺ + 0.025M Na⁺ + 4.045M Clˉ

=> 0.025M NH₄OH + 4.02M NH₄⁺ + 0.025M Na⁺ + 4.045M Clˉ

∴0.025M NH₄OH + 4.02M NH₄⁺ is the buffer solution. 0.025M Na⁺ and 4.045M Clˉ are not reactive.

pH of buffer solution:

NH₄OH ⇄ NH₄⁺ + OHˉ

C(i) 0.025M 4.02M 0M

ΔC -x +x +x

C(eq) (0.025-x)M (4.02+x)M x

≅ 0.025M ≅ 4.02M

Kb = [NH₄⁺][OHˉ]/[NH₄OH] => [OHˉ] = Kb[NH₄OH]/[NH₄⁺]

[OHˉ] = [(1.8 x 10ˉ⁵)(0.025)/(4.02)]M = 1.12 x 10ˉ⁷

=> pOH = -log[OHˉ] = -log(1.12 x 10ˉ⁷) = 6.95

=> pH = 14 – pOH = 14 – 6.95 =7.04 (buffer solution is neutral)

pH of buffer solution after adding 3ml of 0.034M HCl:

… moles HCl added = 0.003L(0.034M) = 1.02 x 10ˉ⁴mole

… Molarity HCl added = 1.02 x 10ˉ⁴mole/(25 + 3)ml = 1.02 x 10ˉ⁴/0.028L =3.643 x 10ˉ³M HCl

NH₄OH => NH₄⁺ + OHˉ

C(i) 0.025M 4.02M 1.12 x 10ˉ⁷M ~ (0)M*

ΔC -3.643 x 10ˉ³M +3.643 x 10ˉ³M +x

C(eq) 0.0214M 4.0236M x**

*Starting concentration of OHˉ is negligible and is assumed to be zero.

** concentration of OHˉ after adding HCl (expect a more acidic system).

[OHˉ](new) = Kb[NH₄OH]/[NH₄⁺]

[OHˉ] = [(1.8 x 10ˉ⁵)(0.0124)/(4.0236)]M = 5.55 x 10ˉ⁸M

=> pOH = -log[OHˉ] = -log(5.55 x 10ˉ⁸) = 7.26

=> pH = 14 – pOH = 14 – 7.26 =6.74 (solution is slightly acidic after adding acid) => pH shifts from 7.04 to 6.74 upon addition of 3ml(0.034M HCl).

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A5.36–g sample of nh4cl was added to 25.0 ml of 1.00 m naoh and the resulting solution diluted to 0....

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