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# Acids and bases - concentration help! watch

1. Calculate the pH of the buffer solution formed when 10.00 cm3 of
0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3 ethanoic acid. Ka, for ethanoic acid in aqueous solution is 1.74 × 10–5 mol dm–3

Mol OH– = (10.0×10-3) ×0.10 =1.0×10-3
Orig mol HA = (25.0 × 10-3) × 0.41 = 0.01025

mol HA in buffer = orig mol HA–mol OH– = 0.00925

Why, in this question do we not convert 1.0x10-3 and 0.00925 into concentrations when we put it in the Ka expression, since they're [ ]
Really confused
2. (Original post by xDesertRose)
Calculate the pH of the buffer solution formed when 10.00 cm3 of
0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3 ethanoic acid. Ka, for ethanoic acid in aqueous solution is 1.74 × 10–5 mol dm–3

Mol OH– = (10.0×10-3) ×0.10 =1.0×10-3
Orig mol HA = (25.0 × 10-3) × 0.41 = 0.01025

mol HA in buffer = orig mol HA–mol OH– = 0.00925

Why, in this question do we not convert 1.0x10-3 and 0.00925 into concentrations when we put it in the Ka expression, since they're [ ]
Really confused
You can if you want, but it makes no difference.

The definition of concentration is moles/volume

So if you have:

(moles salt/volume)/(moles acid/volume)

and both of the volumes are the same (as they have to be as its the same solution)

Then the volumes can be cancelled out to give:

(moles salt/volume)/(moles acid/volume) = (moles salt)/(moles acid)
3. (Original post by charco)
You can if you want, but it makes no difference.

The definition of concentration is moles/volume

So if you have:

(moles salt/volume)/(moles acid/volume)

and both of the volumes are the same (as they have to be as its the same solution)

Then the volumes can be cancelled out to give:

(moles salt/volume)/(moles acid/volume) = (moles salt)/(moles acid)
Ohh ok, but you see the 0.00925 I converted it into conc. by using the overall volume; 0.00925x1000/35 = 0.264

So then 1.74x10-5 x 0.264 / 0.1 = 4.59x10-5 = [H+]

Therefore pH = 4.59x10-5 which is not the answer :/
4. (Original post by xDesertRose)
Ohh ok, but you see the 0.00925 I converted it into conc. by using the overall volume; 0.00925x1000/35 = 0.264

So then 1.74x10-5 x 0.264 / 0.1 = 4.59x10-5 = [H+]

Therefore pH = 4.59x10-5 which is not the answer :/
The first stage is always to calculate the excess and limiting reagent to find the moles of each component in the mixture

Calculate the pH of the buffer solution formed when 10.00 cm3 of
0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3 ethanoic acid. Ka, for ethanoic acid in aqueous solution is 1.74 × 10–5 mol dm–3
mol KOH = 0.01 * 0.1 = 0.001 mol
mol ethanoic acid = 0.025 * 0.41 = 0.01025

As it's a 1:1 reaction the KOH is the limiting reagent and all reacts forming the same number of moles of salt = 0.001

Moles of ethanoic acid remaining = 0.01025 - 0.001 = 0.00925

ka = [H+][A-]/[HA]

[H+] = ka[HA]/[A-] = 1.74 x 10-5 * 0.00925/0.001

[H+] = 1.74 x 10-5 * 9.25 = 0.000161

pH = 3.79
5. (Original post by charco)
The first stage is always to calculate the excess and limiting reagent to find the moles of each component in the mixture

mol KOH = 0.01 * 0.1 = 0.001 mol
mol ethanoic acid = 0.025 * 0.41 = 0.01025

As it's a 1:1 reaction the KOH is the limiting reagent and all reacts forming the same number of moles of salt = 0.001

Moles of ethanoic acid remaining = 0.01025 - 0.001 = 0.00925

ka = [H+][A-]/[HA]

[H+] = ka[HA]/[A-] = 1.74 x 10-5 * 0.00925/0.001

[H+] = 1.74 x 10-5 * 9.25 = 0.000161

pH = 3.79
Thanks, makes sense

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Updated: January 2, 2013
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