Question

What is the equilibrium pH of an initially 4.4 M solution of hypochlorous acid, HOCl, at 25°C (Ka = )? What is Ka for a weak monoprotic acid if a 0.020 M solution of the acid has a pH of 3.28 at 25°C? 15.0 mL of 0.50 M HCl is added to a 100.0-mL sample of 0.484 M HNO2 (Ka for HNO2 = 4.0 × 10–4). What is the equilibrium concentration of NO2– ions?

Answer #1

**What is the equilibrium pH of an initially 4.4 M
solution of hypochlorous acid, HOCl, at 25°C (Ka = )?**

HClO = 4*10^-8

First, assume the acid:

HF

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 4.4 M; then

x^2 + (4*10^-8)x - 4.4*(4*10^-8) = 0

solve for x

x =4.195*10^-4

substitute

[H+] = 0 + 4.195*10^-4= 4.195*10^-4 M

pH = -log(H+) = -log(4.195*10^-4) = 3.377

pH = 3.377

What is the pH of a solution that is 0.30M HOCl (hypochlorous
acid) and 0.15M NaOCl after 0.070 mol HCl/L has been bubbled into
the solution? Ka (HOCl) = 3.5x10-8
HOCl(aq) + H2O <--->
H3O+(aq) +
OCl-(aq)

15.0 mL of 0.50 M HCl is added to a 100.-mL sample of 0.493 M
HNO2 (Ka for HNO2 = 4.0 × 10–4). What is the equilibrium
concentration of NO2 – ions?

Calculate the pH of a 0.375 M aqueous solution
of nitrous acid (HNO2,
Ka = 4.5×10-4) and the
equilibrium concentrations of the weak acid and its conjugate
base.
pH
=
[HNO2 ]equilibrium
=
M
[NO2-
]equilibrium
=
M

a. Find the pH of a 0.100 M solution of a weak
monoprotic acid having Ka= 1.1×10−5.
b.Find the percent dissociation of this solution.
c. Find the pH of a 0.100 M solution of a weak
monoprotic acid having Ka= 1.8×10−3
d. Find the percent
dissociation of this solution.
e.Find the
pH of a 0.100 M solution of a weak monoprotic
acid having Ka= 0.16.
f.Find the percent dissociation of this solution.

If the Ka of a monoprotic weak acid is 4.7 × 10-6, what is the
pH of a 0.50 M solution of this acid?

A 0.50 M solution of a weak acid HA has the same pH as a 0.075 M
solution of HCl. Calculate the Ka for HA

A) HClO is a weak acid (Ka = 4.0 × 10–8) and so the salt NaClO
acts as a weak base. What is the pH of a solution that is 0.087 M
in NaClO at 25 °C? The hint is: Na is a spectator ion in this
situation. Start by finding the Kb of ClO–.
B) A solution of household bleach contains 5.25% sodium
hypochlorite, NaOCl, by mass. Assuming that the density of bleach
is the same as water, calculate...

. A solution of hypochlorous acid (pKa = 7.53) with an initial
volume of 50.0 mL and a formal concentration of 0.120 M is titrated
with 0.120 M potassium hydroxide. Complete the table below by
calculating the pH of the hypochlorous acid solution after the
specified volumes of titrant are added. Then, circle the principle
species. If both species are present at an equal concentration,
circle both. Show all work on separate sheet(s). Volume of titrant
added (mL) pH Principle...

The pKa of hypochlorous acid is 7.530. A 54.0 mL solution of
0.147 M sodium hypochlorite (NaOCl) is titrated with 0.253 M HCl.
Calculate the pH of the solution
a) after the addition of 12.5 mL of 0.253 M HCl.
pH= ?
b) after the addition of 33.5 mL of 0.253 M HCl.
pH=?
c) at the equivalence point with 0.253 M HCl.
pH=?

The pH of a 100.0 mL 0.125 M solution of HA is measured to be
4.1. Calculate Ka for this monoprotic acid.

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