(1)A 100.0 mL sample of 0.180 M HClO4 is titrated with 0.270 M LiOH. Determine the pH of the solution after the addition of 75.0 mL of LiOH.
(2) A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The Ka of HF is 3.5 x 10^-4
please show me the steps. thank you so much
(2) A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The Ka of HF is 3.5 x 10^-4
please show me the steps. thank you so much
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1)A 100.0 mL sample of 0.180 M HClO4 is titrated with 0.270 M LiOH. Determine the pH of the solution after the addition of 75.0 mL of LiOH.
http://en.wikipedia.org/wiki/Perchloric_…
pKa = -8
Ka = 10^8
So HClO4 is a strong acid
HClO4 + LiOH → LiClO4 + H2O
1 mole of HClO4 reacts with 1 mole of LiOH to produce 1 mole of LiClO4 + 1 mole of H2O
Moles of HClO4 = 0.100 liters * 0.180 mole/Liter = 0.018 mole
Moles of LiOH = 0.075liters * 0.270 mole/ Liter = 0.02025
0.018 mole of HClO4 will react with 0.018 mole of LiOH, to produce 0.018 of LiClO4 and 0.018 mole of H2O
Moles of LiOH remaining = 0.02025 – 0.018 = 0.00225 moles of LiOH
The final solution contains 0.018 mole of LiClO4 and 0.00225 mole of LiOH
Total volume = 100 + 75 = 175ml = 0.175 liters
Concentration of OH-1 ions = 0.00225 ÷ 0.175 = 0.0129 = [OH-1]
[H3O+1] * [OH-1] = 10^-14
[H3O+1] = 10^-14 ÷ 0.0129 = 7.75 * 10^-13
pH = -1 * log 7.75 * 10^-13 = 12.11
(2) A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The Ka of HF is 3.5 x 10^-4
ph = pKa + log [A-]/[HA]
HF + KOH → KF + H2O
moles of HF = 0.100 * 0.2 = 0.02
moles of KOH = 0.100 * 0.1 = 0.01
0.01 mole of KOH reacts with 0.01 mole of HF, producing 0.01 mole of KF and 0.01 mole of H2O
AND
0.02 – 0.01 = 0.01 mole of HF remaining
Final solution contains 0.01 mole of KF and 0.01 mole of HF
Total volume = 200 ml = 0.2 liters
[HF] = 0.01÷ 0.2 = 0.05
[F-] = 0.01÷ 0.2 = 0.05
ph = pKa + log [A-]/[HA]
pKa = -1 * log Ka = -1 * log 3.5 * 10^-4 = 3.456
pH = 3.456 + log (0.05/0.05) = 3.456 + 0
pH = 3.456
http://en.wikipedia.org/wiki/Perchloric_…
pKa = -8
Ka = 10^8
So HClO4 is a strong acid
HClO4 + LiOH → LiClO4 + H2O
1 mole of HClO4 reacts with 1 mole of LiOH to produce 1 mole of LiClO4 + 1 mole of H2O
Moles of HClO4 = 0.100 liters * 0.180 mole/Liter = 0.018 mole
Moles of LiOH = 0.075liters * 0.270 mole/ Liter = 0.02025
0.018 mole of HClO4 will react with 0.018 mole of LiOH, to produce 0.018 of LiClO4 and 0.018 mole of H2O
Moles of LiOH remaining = 0.02025 – 0.018 = 0.00225 moles of LiOH
The final solution contains 0.018 mole of LiClO4 and 0.00225 mole of LiOH
Total volume = 100 + 75 = 175ml = 0.175 liters
Concentration of OH-1 ions = 0.00225 ÷ 0.175 = 0.0129 = [OH-1]
[H3O+1] * [OH-1] = 10^-14
[H3O+1] = 10^-14 ÷ 0.0129 = 7.75 * 10^-13
pH = -1 * log 7.75 * 10^-13 = 12.11
(2) A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The Ka of HF is 3.5 x 10^-4
ph = pKa + log [A-]/[HA]
HF + KOH → KF + H2O
moles of HF = 0.100 * 0.2 = 0.02
moles of KOH = 0.100 * 0.1 = 0.01
0.01 mole of KOH reacts with 0.01 mole of HF, producing 0.01 mole of KF and 0.01 mole of H2O
AND
0.02 – 0.01 = 0.01 mole of HF remaining
Final solution contains 0.01 mole of KF and 0.01 mole of HF
Total volume = 200 ml = 0.2 liters
[HF] = 0.01÷ 0.2 = 0.05
[F-] = 0.01÷ 0.2 = 0.05
ph = pKa + log [A-]/[HA]
pKa = -1 * log Ka = -1 * log 3.5 * 10^-4 = 3.456
pH = 3.456 + log (0.05/0.05) = 3.456 + 0
pH = 3.456
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(1) Set up the balanced equation for the reaction. We can assume that both the acid and base are strong, so the reaction will go to completion.
Using that, find out the moles of each acid and base used, and then, using the balanced equation, find out how much of H+ or OH- is left over. Divide by total volume to get the concentration of H+ or OH-.
(2) This is a buffer solution. Use the Henderson-Hasselbach equation. Adding strong base to the acid will produce the conjugate base of the weak acid.
Using that, find out the moles of each acid and base used, and then, using the balanced equation, find out how much of H+ or OH- is left over. Divide by total volume to get the concentration of H+ or OH-.
(2) This is a buffer solution. Use the Henderson-Hasselbach equation. Adding strong base to the acid will produce the conjugate base of the weak acid.