Question

Need help with a titration task (attached file). How should I best approach this problem?

Answer 1

First write the balanced chemical equation
Since we are given with the concentration and volume of sulfuric apply C=n/V and find the no. of moles (n)
According to the stoichiometric ratios find the no. of moles of ammonia.
And then apply C=n/V and find the concentration (C)

Answer 2

H2SO4 + 2NH3 --> (NH4)2SO4
According to the equation we can see the no. of moles ammonia is twice the no. of moles of sulfuric . So find the no. of moles of sulfuric and multiply it by 2 to find the no. of moles of ammonia.

Answer 3

Thank you for your response @Rushwr! It sounds like what I first did, but still, I didn't manage to get the correct answer. Could you help me see where I'm wrong?

How I calculated it:

2NH3 + H2SO4 --> (NH4)2SO4
n(H2SO4) = (0.150 mol dm^-3)/(0.0201 dm^3) = 7.46268... mol
n(NH3) = 2*n(H2SO4) = 2*7.46268... = 14.92537... mol
c(NH3) = nv = 14.92537... * 0.025 = 0.373 mol dm^-3

The mark scheme says that the concentration of NH3 is 0.241 mol dm^-3.

Again; thank you for your response!

Answer 4

There's a calculation error in the moles of sulfuric.

Answer 5

Actually you don't have to convert the volumes to dm^3 cuz it cancels off in the final equation.
Let me show you.
C=n/V right?
And we know no. of moles of ammonia is twice of sulfuric.
No. of moles of sulfuric = n
\[n= 0.15moldm ^{-3} * 20.1cm ^{3}\]
no. of moles of ammonia is 2n right?
So \[2n = 0.15moldm ^{-3}* 20.1cm ^{3} * 2\]
Now apply the values for C=n/V
\[C= \frac{ 0.15moldm ^{-3} * 20.1cm ^{3} * 2 }{ 25cm ^{3} }\]
Since cm^3 is cancelled off we don't actually have to solve that.
so C= 0.241moldm^-3

Answer 6

Ahhh, now I see! I was a little too fast when thinking about that c = n/v formula and did n = c/v instead. And now I realize that it was unnecessary to convert dm^3 to cm^3 as you said. THANK YOU SO MUCH!