[T]

Hello,

The question is: Trusty Dusty is up to his old tricks again – poisoning his mother’s pot plants with aqueous acid-laced fertiliser. Realising that the acid (HX – molar mass 27.03 g mol^–1) is potentially incredibly dangerous to humans (quite volatile) he realises that he needs the free anion (X^–) concentration in his fertiliser to be 0.00340 mol L^–1. First, he does a quick test. If, when he adds the equivalent of 0.560 g of HX to 50.0 mL of water, the pH is measured to be 3.67, how much HX should he add to 4.00 L in order to prepare his fatal fertiliser?

So let y=mol of H⁺ in the solution with pH 3.67

-log₁₀(y/0.05)=3.67
10^-3.67=y/0.05
y=1.069×10^-5

so from this we know that from 0.56 g of HX there will be 1.069×10^-5 of H⁺ and X^- produced.

let z = mol of X^- in 0.0034 molarity 4L solution

z/4=0.0034
z=0.0136
mol of z = mol of H⁺ in this solution.

so therefore dividing 0.0136 by 1.069×10^-5 will tell me how many 0.56g of HX is needed. 0.0136/1.069×10^-5  = 1272.24 times of 0.56 g needed.
1272.24 × 0.56 = 712.45 g of HX. However this is not the same as the answer given. Could someone tell me what I did wrong?

Thanks

[Dan]

So let y=mol of H⁺ in the solution with pH 3.67

-log₁₀(y/0.05)=3.67
10^-3.67=y/0.05
y=1.069×10^-5

so from this we know that from 0.56 g of HX there will be 1.069×10^-5 of H⁺ and X^- produced.

I think you are ok up to here.

From a cursory glance it looks like from here you have based you subsequent calculations on the assumption that the concentration [X^-] is linearly related to the amount of HX added. This not true for a weak acid, consider:

Ka = [H⁺][X^-]/[HX]

You have enough information to calculate Ka - do that and g from there.


--
*Demonstration of non linearity:
 
Let [H⁺] = [X^-] = x; [HX] = C, where C is the initial concentration of HX (before equilibrium is reached)

Ka = x²/(C-x)
x² + xKa - CKa = 0
x = -Ka ± sqrt(Ka² + 4CKa)

This shows that, for example, doubling C does not double x. The relationship is non linear.

[T]

I see thanks Dan.

K_a=((2.13×10^-4))/(0.0207/0.05-2.13×10^-4)
Ka = 1.10×10^-7

We know the molarity of X=0.00340 in 4 L solution
So if amount of mol of HX added to 4 L solution = y

1.10×10^-7 = (0.00340²)/([y/4]-0.0034)
y/4-0.00340 = (0.00340²)/(1.10×10^-7)
y=4.186×10²

4.186×10² × 27.03 = 11.3 kg.

Thanks Dan


 

[Dan]

Is 11.3 kg in 4 L the correct answer? That is the value I'm getting as well, but it seems wildly unrealistic. That's a 100 M solution...

[Dan]

when he adds the equivalent of 0.560 g of HX to 50.0 mL of water, the pH is measured to be 3.67

I think this is where the mistake is - he adds an amount of HX to 50 mL. The amount is the equivalent of 0.56 g in 4 L - i.e. 0.007 g in 50 mL.

Carrying that through gives a much more sensible value.

[T]

Hi Dan, 11.3 kg was the correct answer. I agree that that is a lot of HX to add to a 4L solution though.