Question

Determine the pH of an HF solution of each of the following concentrations. 0.270 M, 4.90×10−2 M, 2.30×10−2 M

Answer #1

HF is a weak acid

so

HF <-> H+ + F-

Ka = [H+][F-]/[HF]

for HF = 7.2*10^-4

so

[H+] = [F-] = x in equilibrium

[HF] = M-x

substituting inf Ka

Ka = x*x/(M-x)

7.2*10^-4 = x^2 / (M-x)

x^2 + Ka*x -MKa = 0

solve for each

M = 0.27

x^2 + Ka*x -**0.27***7.2*10^-4 = 0

x = H+ = 0.0135

ph = -log(0.0135) = 1.8696

x^2 + Ka*x -**4.9*10^-2***7.2*10^-4 = 0

x = H+ = 0.00559

pH = -log(0.00559) = 2.2525

x^2 + Ka*x - **2.3*10^-2***7.2*10^-4 = 0

x = H+ = 0.0037

pH = -log(0.0037) = 2.4317

PART A:
Determine the pH of an HF solution of each of the following
concentrations.
0.260M
4.70×10−2 M
2.30×10−2 M
In which cases can you not make the simplifying assumption that
x is small? (choose one of the below options)
only in (a)
only in (b)
in (a) and (b)
in (b) and (c)
PART B:
A 8.0×10−2 M solution of a monoprotic acid has a
percent dissociation of 0.60%. Determine the acid ionization
constant (Ka) for the acid.

Determine the pH of an HF solution of each of the following
concentrations. a. .290 M b. 5.20 x 10^-2 M c. 2.10 x 10^-2 M d. in
which cases can you not make simplifying assumption that x i
small

Determine the pH of an HF solution of each of the following
concentrations. In which cases can you not make the simplifying
assumption that x is small? (Ka for HF is 6.8×10−4.) a) 0.300 M b)
4.70×10−2 M c) 2.60×10−2 M
d) In which cases can you not make the simplifying assumption
that x is small?
only in (a)
only in (b)
in (a) and (b)
in (b) and (c)

Determine the pH of an HF solution of each of the following
concentrations. In which cases can you not make the simplifying
assumption that x issmall?
(Ka for HF is 6.8×10−4.)
Part A
0.260 M
Part B
5.30×10−2 M
Part C
2.60×10−2 M
Express your answers to two decimal places.
In which cases can you not make the simplifying assumption that
x is small?
only in (a)
only in (b)
in (a) and (b)
in (b) and (c)

What is the pH of a 0.010 M solution of HF? (The Ka value for HF
is 6.8×10−4.) What is the of a 0.010 solution of ? (The value for
is .)
a. 1.58
b. 2.10
c.2.30
d. 2.58
e. 2.64

Determine the pH of each of the following
two-component solutions.
0.270 M NH4NO3 and 0.101 M HCN
7.5×10−2 M RbOH and 0.130 M NaCl
9.2×10−2 M HClO4 and 2.0×10−2 M KOH
0.110 M NaClO and 5.50×10−2 M KI

Calculate the pH of a solution that contains the following
analytical concentrations:
a. 0.222 M H3PO4 and 0.411 M
NaH2PO4
pH=
b. 0.0270 M in Na2SO3 and 0.0311 M in
NaHSO3
pH=
c. 0.270 M in HOC2H4NH2 and
0.180 M in HOC2H4NH3Cl
pH=
d. 0.0130 M in H2C2O4 and
0.0400 M in Na2C2O4
pH=
e. 0.0270 M in Na2C2O4 and
0.0180 M in NaHC2O4
pH=

Find the pH of a 0.100 M HF solution. Find the percent
dissociation of a 0.100 M HF solution Find the pH of a 6.00×10?2 M
HF solution. Find the percent dissociation of a 6.00×10?2 M HF
solution.

Determine the pH of each solution.
8.40×10−2 M HClO4
a solution that is 5.2×10−2 M in HClO4 and
4.7×10−2 Min HCl
a solution that is 1.06% HCl by mass (Assume a density of 1.01
g/mL for the solution.)

A 100mL sample of.20 M HF is titrated with 0.10 M KOH. Determine
the pH of the solution after the addition of 60.0 mL of KOH. Ka
(HF) = 3.5*10^-4

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