0.1 of concentration(0.1M) of NaOH in the buret and there is 10 ml unknown acid

Question

0.1 of concentration(0.1M) of NaOH in the buret and there is 10 ml unknown acid in the flask.

I used 14.2 ml of NaOH to neturalize unknown acid.

My question is,

1) What would happen to the concentraion of unknown acid if the flask was contaminated with other types of acid?

2.) what is the concentration of your unknown acid if it was given as diprotic acid?

3) What is the concentraion of unknown acid if the tip of the burette was not filled when reading the initial volume?

4) What is the concentraion of unknown if you used Ca(OH)2 instead of NaOH and did not know it?

Next experiment, I used KHP in the flask instead of unknown acid.

KHP I used is not pure.

So my question is,

What would have happened to the final result if the impurity in the sample was also acidic?

Explanation / Answer

1)

Let he moles of acid reacted with the base be y

now

if the acid is not pure

actual moles of acid reacted will be less than y

but we neglect it

and use the value y in our calculations

so

the calculated conc of the uknown acid is higher than the actual conc of acid

2)

we know that

moles = molarity x volume (L)

so

moles of NaOH used = 0.1 x 14.2 x 10-3

moles of NaOH used = 1.42 x 10-3

now

given a diprotic acid

we know that

2 moles of NaOH reacts with 1 mole of diprotic acid

So

moles of acid = 0.5 x moles of NaOH

moles of acid = 0.5 x 1.42 x 10-3

moles of acid = 7.1 x 10-4

now

given 10 ml of acid

So

molarity = moles /volume(L)

so

molarity of acid = 7.1 x 10-4 / 10 x 10-3

molarity of the acid = 0.071 M

so the conc of unknown acid is 0.071 M

3)

if the tip of the burette was not filled initially

then the actual volume of NaOH used will be less than 14.2 ml

as moles = molarity volume

actual moles of NaOH used will be less than 1.42 x 10-3

now

moles of acid = 0.5 x moles of NaOH used

so

actual moles of acid present will be less than 7.1 x 10-4

now

conc of acid = moles /volume

so

actual conc of acid will be less than 0.071 M

finally

by not filling the tip of the burette before taking the initial reading

we get a concentration of the acid which is higher than the actual value

4)

given Ca(OH)2 is used instead of NaOH

as

1 mole of Ca(OH)2 reacts with 1 mole of diprotic acid

moles of acid = moles of Ca(OH)2 used

moles of acid = 1.42 x 10-3

now

conc of acid = 1.42 x 10-3 / 10 x 10-3

conc of acid = 0.142 M

So if Ca(OH)2 is used instead of NaOH

the concentration of unknown acid will be 0.142 M

Given KHP is impure

So the calculated conc of KHP will be higher than the actual conc of KHP

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