Question

Hey guys, I am having a lot of trouble with my homework. Could you please help? I thumbs up anyone who gives a thorough explanation and shows the work! I really need to understand these concepts!

1) At 25 °C, the pH of a 0.15 M aqueous solution of CsZ (the cesium salt of HZ) is 10.70. What is the Ka for HZ? 2) Calculate the pH of 0.586 M anilinium hydrochloride (C6H5NH3CI) solution in water at 25 °C, given that Kb for aniline is 3.83 x 10-4. 3) Kb for NH3 is 1.8 x 10-5. What is the pH of a 0.45 M aqueous solution of NH4Cl at 25.0 °C?

Thank you!

Answer 1

1)

given, pH = 10.70 $\Rightarrow$ pOH = 14 - pH = 3.30 $\Rightarrow$ [OH^-] = 5.01 $\times$ 10^-4 M.

the reaction is CsZ + H₂O  $\rightleftharpoons$  Cs⁺ + OH^- + HZ

initial concentrations: 0.15 0 0 0

final concentrations: 0.15-x x x x

x =  5.01 $\times$ 10^-4 M << 0.15M.

Therefore, 0.15-x = 0.15(approximately).

Now, [Z^-](which is nothing but the [CsZ] because CsZ exists as Cs⁺ and Z^- in water) = 0.15 <.

[HZ] = x = 5.01 $\times$ 10^-4 M.

[H⁺] = 2 $\times$ 10^-11 M

The dissociation reaction of acid HZ is: HZ $\rightleftharpoons$    H⁺ + Z^-

Ka = [H⁺][Z^-]$\div$[HZ] = ( 2 $\times$ 10^-11 M ) $\times$ (0.15 M ) $\div$ ( 5.01 $\times$ 10^-4 M ) = 6.0  $\times$ 10^-9 M

So, we just have to find the concentration and us appropriate equation to find the dissociation constants.

We can only answer one question.