5. Accurately determine the pH of 0.0995 M sulfuric acid. Ka1 >> 1 and Ka2 =0.01

Question

5. Accurately determine the pH of 0.0995 M sulfuric acid. Ka1 >> 1 and Ka2 =0.012.

9. The equilibrium constant, Ka, for the reaction given below is is 6.0 x 10-3. Calculate the pH of a 0.798 M solution of Fe(H20)63+.

Fe(H20)63+(aq) + H2O(l) <--> Fe(H20)5OH2+(aq) + H3O+(aq)

10. Calculate the pH necessary for 97.4 % of the iron(III) to be in the form Fe(H20)63+

I'm having a bit of trouble with these particular problems. Thank you!

5. Accurately determine the pH of 0.0995 M sulfuric acid. Ka1 > > 1 and Ka2 =0.012. 9. The equilibrium constant, Ka, for the reaction given below is is 6.0 x 10-3. Calculate the pH of a 0.798 M solution of Fe(H20)63+. Fe(H20)63+(aq) + H2O(l) Fe(H20)5OH2+(aq) + H3O+(aq) 10. Calculate the pH necessary for 97.4 % of the iron(III) to be in the form Fe(H20)63+ I'm having a bit of trouble with these particular problems. Thank you!

Explanation / Answer

5 -

H2SO4 >> H+ + HSO4-
this gives us
9.95 x 10^-2 M of H+ and 2.00 x 10^-2 M of HSO4-

HSO4- <-----> H+ + SO42-
1.2 x 10^-2 = ( 9.95 x 10^-2+x) (x) / 2.00 x 10^-2 -x

solve for x by quadratic formula and remember that [H+] = x + 2.00 x 10^-2

9 -

[H+] = square root (Ka * Ca) = square root (6.0 x 10-3 * 0.798) = 4.78*10^-3
pH = - log( 4.78*10^-3 ) = 2.32

10 -

it is also like 9 so i think with the help of 9 you can solve it

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