Can you please help me these?

Question

The [H₃O⁺] in a solution of 2.09 mol/L benzoic acid in water is a.bcdef mol/L

The [OH^-] of a solution of 3.77 mol/L sodium carbonate is a.bcd mol/L

The pH of a solution of 1.13 mol/L benzoic acid in water is a.bc

The pH of a 0.401 mol/L solution of sulfurous acid is 1.16. Based on this evidence, the equilibrium concentration of H₂SO_3(aq) is a.bc mol/L.

Answer 1

Is this some kind of bizarro number puzzle where you are solving for the digits a-f? If so, I like it! If the letters are only indicating where things are rounded, you'd need the K_A of benzoic acid. Let's assume the weird one...

The ICE calculation for the K expression given an acid is always the same for a B/L acid.

From statement (4) : We can find [H⁺] from pH.: [H⁺] = 10^-pH = x

If we assume that sulfurous acid is the sole source of H⁺ and HSO₃^-,

K_A (SA) = x²/(.401 - x) (answer to 4 and you know a.bc

Using statement (3) We can find the [H⁺] concentration by taking 10^-a.bc = y where a.bc is the pH

K_A (BA) = y²/(1.13 - y)

We can use the K from (3) to solve for the molarity of H⁺ (let's call it z) in (1):

K_A (BA) = z² / ( 2.09 - z) (you probably have to solve the quadratic for z to get all those places.

I'll leave (2) for you. You can use it to check d or check that you can do this.

Good luck!

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