Question

Determine the pH of a 0.034 solution of H2SO4. The
dissociation occurs in two steps. Ka1 is extremely large; Ka2 is
1.2 * 10^-2

Please show work thanks

Answer #1

_{a1} is large so the first proton is fully
dissociated

This gives [H^{+}] of 3.4 x10^{-2} = 0.034

The second proton is weakly dissociated so

K_{a2} =
[H^{+}][SO_{4}^{2-}]/[HSO_{4}^{-})]
= 1.2x10^{-2} = 0.012

We know that the total concentration of HSO_{4}^{-}
present is the same as the initial concetration of
H_{2}SO_{4} because the first dissociation is
complete.

Let the amount of dissociation of HSO_{4}^{-} be
x.

After dissociation [HSO_{4}^{-} ]= (0.0032 - x) and
[H^{+}] = [SO_{4}^{2-}] = x

But from K_{a1} we already have [H^{+}] of 0.0032
so total [H^{+}] = 0.0032 + x

From K_{a2}:

0.012 = (0.0032 + x) x/(0.0032 - x)

0.012 (0.0032 - x) = (0.0032 + x) x

0.0000384 - 0.012x = 0.0032x + x^{2}

x^{2} + 0.0152x - 0.0000384 = 0

Solve this quadratic for x using the quadratic formula.

on calculation we get, x = 0.0022

This is the concentration of H^{+} from the second
dissociation

Total [H^{+}] = 0.0032 + 0.0022 = 0.0054

pH = - log(.0054) = 2.27

Calculate the pH of 0.15 M H2SO4(aq) at 25◦C, given that Ka1 is
very large and Ka2 = 1.2 × 10−2

What is the pH of a 0.280 M solution oof
H2SO4? Ka2 =1.2 x
10-2
Please help and show work, thank you

Calculate the [C3H2O42-] and the pH of a 0.10 M solution of
malonic acid (H2C3H2O4) (Ka1 = 1.5 x 10-3 ; Ka2 = 2.0 x 10-6).
Is there a way to do this without the ICE tables? Please show
work. Thanks!

For which .10M diprotic acid would the 2nd dissociation affect
the pH significantly? Explain.
H2A: Ka1=4.2*10^-2 Ka2=1.8*10^-7
H2B: Ka1=2.4*10^-4 Ka2= 6.1*10^-8
H2C: Ka1=1.3*10^-4 Ka2=5.2*10^-9
H2D: Ka1=1.8*10-3 Ka2=9.3*10^-4

What is the pH of a 0.260 M solution of H2SO4? Ka2 =
1.20×10^–2?
Please write out steps if possible so that I can understand,
thank you!

What is the pH of a 0.170 M solution of sulfurous acid? Given:
Ka1 = 1.70×10–2, Ka2 = 6.20×10–8 Please show the setup of the
quadratic if possible. Thank you in advance!

what is the pH of a 0.100M soln of H2SO4? Given Ka1=
1.70x10-2, Ka2=6.20x10-8
answer is not 1.39.

What is the pH of a solution of
3.4 M H2A (Ka1 = 1.0 × 10^-6 and Ka2 is 1.0 × 10^-10)?
Please explain step by step

2. Consider the triprotic acid, H3A, with the
following acid dissociation contants: Ka1
= 1.5 x 10-5 ,
Ka2= 5.0 x 10-9 , and Ka3 = 5.0 x
10-12
a.) What is the pH of an
aqueous solution of 0.200 M H3A?
b.) What is the pH of an aqueous solution of 0.200 M
NaH2A?
c.) What is the pH of an aqueous solution of 0.200 M
Na2HA?
d.) What is the pH of an aqueous solution of 0.200 M...

1. What is the pH of a 1.05M solution of carbonic acid? Ka1 =
4.3 x 10-7 Ka2 = 5.6 x 10-11
2. A diprotic acid, H2A has the following
Kas: Ka1 = 3.94 x 10-7 and
Ka2 = 8.78 x 10-11. What is the Kb
of HA-?
3.A diprotic acid, H2A has the following
Kas: Ka1 = 3.94 x 10-7 and
Ka2 = 8.78 x 10-11. What is the Kb
of HA-?

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