Question

If a solution is 0.10 M phosphoric acid, what is the solution pH? Ka1 = 0.0075.

Answer #1

H3PO4 dissociates as:

H3PO4 -----> H+ + H2PO4-

0.1 0 0

0.1-x x x

Ka = [H+][H2PO4-]/[H3PO4]

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 ((7.5*10^-3)*0.1) = 2.739*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

7.5*10^-3 = x^2/(0.1-x)

7.5*10^-4 - 7.5*10^-3 *x = x^2

x^2 + 7.5*10^-3 *x-7.5*10^-4 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 7.5*10^-3

c = -7.5*10^-4

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.056*10^-3

roots are :

x = 2.389*10^-2 and x = -3.139*10^-2

since x can't be negative, the possible value of x is

x = 2.389*10^-2

So, [H+] = x = 2.389*10^-2 M

use:

pH = -log [H+]

= -log (2.389*10^-2)

= 1.6218

Answer: 1.62

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]

What is the [H2PO4-] of a
solution labeled "0.10 M Phosphoric Acid
(H3PO4)"?
[Ka1 = 7.1x10-3; Ka2 =
6.3x10-8; Ka3 = 4.2x10-13]
A. 4.2x10-13 M
B. 2.7x10-2 M
C. 7.1x10-3 M
D. 1.6x10-9 M
E. 1.6x10-16 M

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!

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.

An aqueous solution of phosphoric acid, that is
2.00 percent by weight phosphoric
acid, has a density of 1.0092 g/mL. A
student determines that the freezing point of this solution is
-0.464 °C.
Based on the observed freezing point, what is the percent
ionization of the acid and the value of
Ka1? Assume that only the first
ionization of the acid is important.
Kf for H2O is 1.86 °C/m.
% ionized
= %
Ka1
=

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

The pH of a 1.0 L solution of 0.10 M lactic acid is 4.34. What
is the new pH of the solution if 40 mL of 1.0 M HCl is added? The
pKa of lactic acid is 3.86.

Show how phosphoric, H3PO4, dissociates in water and calculate
the pH of a 0.030 M solution of the acid. Ka1 = 7.5 x 10-3; K a2 =
6.2 x 10-8; Ka3 = 4.8 x 10-13

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 solution containing 0.10 M sodium acetate to
which is added 0.10 M acetic acid?
What is the pH of a solution prepared by mixing 40 ml of 0.10 M
acetic acid with 60 ml of 1 M sodium acetate?

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 4 minutes ago

asked 6 minutes ago

asked 27 minutes ago

asked 31 minutes ago

asked 45 minutes ago

asked 56 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago