Answer:
A- pH = 13.12
B- pH = 12.91
C- pH = 12.71
D- pH = 12.43
E- pH = 11.55
F- pH = 7
G- pH = 2.46
H- pH = 1.88
Explanation:
This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) → H₂O(l) + KCl(aq)
Our pH at the equivalence point is 7, because we have made a neutral salt.
To determine the volume at that point we state the formula for titration:
mmoles of base = mmoles of acid
Volume of base . M of base = Volume of acid . M of acid
50mL . 0.129M = 0.258 M . Volume of acid
Volume of acid = (50mL . 0.129M) / 0.258 M → 25 mL (Point <u>F</u>)
When we add 25 mL of HCl, our pH will be 7.
A- At 0 mL of acid, we only have base.
KOH → K⁺ + OH⁻
[OH⁻] = 0.129 M
To make more easy the operations we will use, mmol.
mol . 1000 = mmoles → mmoles / mL = M
- log 0.129 = 0.889
14 - 0.889 = 13.12
B- In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺
Initially we have 0.129 M . 50 mL = 6.45 mmoles of OH⁻
1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:
6.45 mmol - 1.81 = 4.64 mmoles of OH⁻
This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.
[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M
- log 0.0815 M = 1.09 → pOH
pH = 14 - pOH → 14 - 1.09 = 12.91
C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺
<em>Our initial mmoles of OH⁻ would not change through all the titration. </em>
Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.
6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻
Total volume is: 50 mL of base + 12.5 mL = 62.5 mL
[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M
- log 0.0517 = 1.29 → pOH
14 - 1.11 = 12.71
D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺
6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.
6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻
Total volume is: 50 mL of base + 18 mL = 68 mL
[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M
- log 0.0265 = 1.57 → pOH
14 - 1.57 = 12.43
E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺
6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.
6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻
Total volume is: 50 mL of base + 24 mL = 74 mL
[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M
- log 3.51×10⁻³ = 2.45 → pOH
14 - 2.45 = 11.55
F- This the equivalence point.
mmoles of OH⁻ = mmoles of H⁺
We add (25 mL . 0.258M) = 6.45 mmoles of H⁺
All the OH⁻ are neutralized.
OH⁻ + H⁺ ⇄ H₂O Kw
[OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ → pOH = 7
pH → 14 - 7 = 7
G- In this case we have an excess of H⁻
We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺
We neutralized all the OH⁻ but some H⁺ remain after the equilibrium
6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺
[H⁺] = 0.26 mmol / Total volume
Total volume is: 50 mL + 26 mL → 76 mL
[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M
- log 3.42×10⁻³ = 2.46 → pH
H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺
We neutralized all the OH⁻ but some H⁺ remain after the equilibrium
7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons
Total volume is 50 mL + 29 mL = 79 mL
[H⁺] = 1.03 mmol / 79 mL → 0.0130 M
- log 0.0130 = 1.88 → pH
After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.
is the pressure of the gas,
is the volume of the gas,
is the number of gas particles in the gas,
is the ideal gas constant, and
is the absolute temperature of the gas in degrees Kelvins.
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