Question

Calculate the concentrations of H, 0* OH, HSO," and Seo, in 0.12MH, Sesselenic acid, solution (This problem requires values in your textbook's specific appendices, which you can access through the OWL 2 Mind Tap Reader. You should not use the OWLV2 References Tables to answer this question as the values will not match.) Concentration of 1,0 - M Concentration of OH- M Concentration of HSO- M Concentration of Seo M

Answer 1

Concentration of H3O+ = 0.13 M

concentration of OH- = 7.7 x 10^-14 M

concentration of HSeO4- = 0.11 M

concentration of SeO4^2- = 0.010 M

First dissociation

H2SeO4 (aq) + H2O (l) $\rightarrow$ HSeO4- (aq) + H3O+ (aq)

Since Ka1 is very large, all H2SeO4 dissociates to HSeO4- and H3O+

now concentration of HSeO4- = initial concentration of H2SeO4 = 0.12 M

concentration of H3O+ = initial concentration of H2SeO4 = 0.12 M

Second dissociation

ICE table	HSeO4- (aq)	H2O (l)	$\rightleftharpoons$	SeO4^2- (aq)	H3O+ (aq)
Initial conc.	0.12 M	-		0	0.12 M
Change	-x	-		+x	+x
Equilibrium conc.	0.12 M - x	-		+x	0.12 M + x

Ka2 = [SeO4^2-]eq[H3O+]eq / [HSeO4-]eq

1.2 x 10^-2 = [(x) * (0.12 M + x)] / (0.12 M - x)

Solving for x, x = 0.010 M

[H3O+] = 0.12 M + x

[H3O+] = 0.12 M + 0.010 M

[H3O+] = 0.13 M