A 25.00mL sample of 0.30 M weak acid (Ka=2.5x10^-6) is titirated with 0.30 M LiO

Question

A 25.00mL sample of 0.30 M weak acid (Ka=2.5x10^-6) is titirated with 0.30 M LiOH. Calaculate the pH at the following points in the titration: a) 6.25 mL of base added b) 12.5 mL of base added c) 18.75 mL of base added

A 25.00mL sample of 0.30 M weak acid (Ka=2.5x10^-6) is titirated with 0.30 M LiOH. Calaculate the pH at the following points in the titration: a) 6.25 mL of base added b) 12.5 mL of base added c) 18.75 mL of base added

a) 6.25 mL of base added b) 12.5 mL of base added c) 18.75 mL of base added

Explanation / Answer

1)when 6.25 mL of LiOH is added

Given:

M(HA) = 0.3 M

V(HA) = 25 mL

M(LiOH) = 0.3 M

V(LiOH) = 6.25 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.3 M * 25 mL = 7.5 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.3 M * 6.25 mL = 1.875 mmol

We have:

mol(HA) = 7.5 mmol

mol(LiOH) = 1.875 mmol

1.875 mmol of both will react

excess HA remaining = 5.625 mmol

Volume of Solution = 25 + 6.25 = 31.25 mL

[HA] = 5.625 mmol/31.25 mL = 0.18M

[A-] = 1.875/31.25 = 0.06M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 2.5*10^-6

pKa = - log (Ka)

= - log(2.5*10^-6)

= 5.602

use:

pH = pKa + log {[conjugate base]/[acid]}

= 5.602+ log {6*10^-2/0.18}

= 5.125

2)when 12.5 mL of LiOH is added

Given:

M(HA) = 0.3 M

V(HA) = 25 mL

M(LiOH) = 0.3 M

V(LiOH) = 12.5 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.3 M * 25 mL = 7.5 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.3 M * 12.5 mL = 3.75 mmol

We have:

mol(HA) = 7.5 mmol

mol(LiOH) = 3.75 mmol

3.75 mmol of both will react

excess HA remaining = 3.75 mmol

Volume of Solution = 25 + 12.5 = 37.5 mL

[HA] = 3.75 mmol/37.5 mL = 0.1M

[A-] = 3.75/37.5 = 0.1M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 2.5*10^-6

pKa = - log (Ka)

= - log(2.5*10^-6)

= 5.602

use:

pH = pKa + log {[conjugate base]/[acid]}

= 5.602+ log {0.1/0.1}

= 5.602

3)when 18.75 mL of LiOH is added

Given:

M(HA) = 0.3 M

V(HA) = 25 mL

M(LiOH) = 0.3 M

V(LiOH) = 18.75 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.3 M * 25 mL = 7.5 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.3 M * 18.75 mL = 5.625 mmol

We have:

mol(HA) = 7.5 mmol

mol(LiOH) = 5.625 mmol

5.625 mmol of both will react

excess HA remaining = 1.875 mmol

Volume of Solution = 25 + 18.75 = 43.75 mL

[HA] = 1.875 mmol/43.75 mL = 0.0429M

[A-] = 5.625/43.75 = 0.1286M

They form acidic buffer

acid is HA

conjugate base is A-

Ka = 2.5*10^-6

pKa = - log (Ka)

= - log(2.5*10^-6)

= 5.602

use:

pH = pKa + log {[conjugate base]/[acid]}

= 5.602+ log {0.1286/4.286*10^-2}

= 6.079

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