Question

Could someone please help me with chemistry of acids and bases.

Answer 1

@aaronq can you please help me. I have been trying so hard to finish this and some questions I try and i keep getting them wrong

Answer 2

Where it is?

Answer 3

Calculate the concentrations of all species in a 1.70 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.

Answer 4

So the ionization given are:

\(H_2SO_3 \rightleftharpoons H^++HSO_3^-\) \(K_{A_1} = 1.4× 10^{–2}\)
\(HSO_3^- \rightleftharpoons H^++SO_3^{2-}\) \(K_{A_2} = 6.3× 10^{–8}\)

but you're working with the conjugate base form, \(SO_3^{2-}\), so you need to convert the \(K_A\)'s given to \(K_B\)'s.

\(K_w=K_A*K_B\rightarrow K_B=\dfrac{K_w}{K_A}\)

so for: \( H_2O+SO_3^{2-}\rightleftharpoons HSO_3^- +OH^- \) \( K_{B_1}=\dfrac{K_w}{K_{A_2}}=\dfrac{1.0*10^{-14}}{ 6.3× 10^{–8}}=1.6*10^{-7}\)

I N/A 1.7M 0 0
C -x +x +x <- ICE table
E 1.7-x x x

now write the equilibrium expression: \( K_{B_1}=\dfrac{[HSO_3^-][OH^-]}{[SO_3^{2-}]}\)

Ignore water because of it's high concentration (55.5 M)

Plug values from ICE table into the expression above and solve: \(1.6*10^{-7}=\dfrac{x^2}{(1.7-x)}\)

\(x=0.0005194 ~M=[OH^-]=[HSO_3^-]\)

\([SO_3^{2-}]=1.7-0.000519383=1.69948~ M\)

you now have to do the next ionization for: \( H_2O+HSO_3^{-}\rightleftharpoons H_2SO_3+OH^- \)

use: \([HSO_3^-]=0.0005194 ~M\) and don't forget to add up your values for \([OH^-]\) for the total.