Respuesta :
Answer:
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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).
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
Therefore, 0.025M NH₄OH + 4.02M NH₄⁺ is the buffer solution. 0.025M Na⁺ and 4.045M Clˉ are not reactive.
What is a buffer solution?
This is an aqueous solution consisting of a mixture of a weak acid and it's conjugate base, or vice versa.
A. 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
B. pH = 14 – pOH = 14 – 6.95 =7.04 (buffer solution is neutral)
C. 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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