Question

a) What is the pH of 0.10 M NaH_{2}PO_{4}?
K_{a2} = 6.2 x 10^{-8}

b) What is the pH of 0.10 M Na_{2}HPO_{4}?
K_{a3} = 2.2 x 10^{-13}

Answer #1

a)

H2PO4- dissociates as:

H2PO4- -----> H+ + HPO42-

0.1 0 0

0.1-x x x

Ka = [H+][HPO42-]/[H2PO4-]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.2*10^-8)*0.1) = 7.874*10^-5

since c is much greater than x, our assumption is correct

so, x = 7.874*10^-5 M

So, [H+] = x = 7.874*10^-5 M

use:

pH = -log [H+]

= -log (7.874*10^-5)

= 4.1038

Answer: 4.10

b)

HPO42- dissociates as:

HPO42- -----> H+ + PO43-

0.1 0 0

0.1-x x x

Ka = [H+][PO43-]/[HPO42-]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.2*10^-13)*0.1) = 1.483*10^-7

since c is much greater than x, our assumption is correct

so, x = 1.483*10^-7 M

So, [H+] = x = 1.483*10^-7 M

use:

pH = -log [H+]

= -log (1.483*10^-7)

= 6.8288

Answer: 6.83

What is the pH of a 0.20 M solution of disodium hydrogen
phosphate? Ka2 = 6.2 x 10^- 8and Ka3 = 2.2 x 10^-1 3

Given that the stepwise dissociation constants for phosphoric
acid are:
Ka1 = 7.5×10-3; Ka2 =
6.2×10-8; Ka3 = 4.8×10-13
To prepare 1.20 L of a buffer solution having an ionic strength
of 0.150 and a pH of 7.50 would require:
Answer to 3 sig figs.
a) (mass) of Na2HPO4(anhydrous) and
b) (mass) of NaH2PO4(anhydrous).

What is the (H2PO4-) of a solution labeled "0.10 M Phosphoric
Acid"? [Ka1= 7.1 x 10 ^ -3 , Ka2= 6.3 x 10^ -8, Ka3 = 4.2 x 10
^-13]

Phosphoric acid is a triprotic acid, and the Ka values are given
below. Calculate pH, pOH, [H3PO4], [H2PO4 2-], [HPO4 -], and [PO4
3-] at equilibrium for a 5.00 M phosphoric acid solution.
Ka1 = 7.5 x 10^-3
Ka2 = 6.2 x 10^-8
Ka3 = 4.2 x 10^-13

Phosphoric acid is a triprotic acid (Ka1 = 6.9× 10–3, Ka2 = 6.2×
10–8, and Ka3 = 4.8× 10–13). To find the pH of a buffer composed of
H2PO4–(aq) and HPO42–(aq), which pKa value would you use in the
Henderson-Hasselbalch equation?
A) pKa1= 2.16
B) pKa2=7.21
C) pKa3= 12.32
Calculate the pH of a buffer solution obtained by dissolving
24.0 g of KH2PO4(s) and 43.0 g of Na2HPO4(s) in water and then
diluting to 1.00 L.

2The acid-dissociation constants of phosphoric acid (H3PO4) are
Ka1 = 7.5 × 10-3, Ka2 = 6.2 × 10-8, and Ka3 = 4.2 × 10-13 at 25.0
°C. What is the molar concentration of phosphate
ion in a 2.5 M aqueous solution of phosphoric acid?
A) 0.13
B) 2.5 × 10-5
C) 8.2 × 10-9
D) 9.1 × 10-5
E) 2.0 × 10-19
Can you include the complete
steps where Ka2 and Ka3 is used.

Calculate the pH of 0.103 M phosphoric acid
(H3PO4, a triprotic acid). Ka1 =
7.5 x 10-3, Ka2 = 6.2 x 10-8, and
Ka3 = 4.8 x 10-13.
Hint, if you are doing much work, you are making the problem harder
than it needs to be.

Calculate the pH and the equilibrium concentrations of H2PO4-,
HPO42- and PO43- in a 0.0353 M aqueous phosphoric acid solution.
For H3PO4, Ka1 = 7.5×10-3, Ka2 = 6.2×10-8, and Ka3 = 3.6×10-13
pH = ?
[H2PO4-] = ?M
[HPO42-] = ?M
[PO43-] = ?M

Calculate the pH during the titration of 50.00 mL of 0.1000 M
phosphoric [H3PO4; Ka1=7.1 x 10^(-3), Ka2=6.3 x 10^(-8), Ka3=4.2 x
10^(-13)] after adding 33.42 mL of 0.1000 M NaOH.

Another buffer found in blood is based on the equilibrium
between dihydrogen phosphate and monohydrogen phosphate. The
reaction is shown below:
H2PO4^-(aq) + H2O(l)----> H3O^+(aq) + HPO4^-2(aq)
If the pH of a blood sample was 7.10, what would you calculate
as the ratio of (H2PO4^-) to (HPO4^-2)?
(Ka1 = 7.5*10^-3, Ka2 = 6.2*10^-8, Ka3 = 3.6*10^-13)

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