Question

Calculate the pH of a 0.39 M CH3COOLi solution. Ka for acetic acid= 1.8 x 10^-5

Answer #1

CH_{3}COOLi is a strong salt, So it dissociates
completely

CH_{3}COOLi
CH_{3}COO^{-} + Li^{+}

Now,

CH_{3}COO^{-} + H_{2}O CH_{3}COOH
+ OH^{-}

Ka for acetic acid= 1.8 x 10^{-5}

Kb = Kw/Ka = 1 x 10^{-14} / 1.8 x 10^{-5} =5.6 x
10^{-10}

But,

Kb = [CH_{3}COOH] [OH^{-} ] /
[CH_{3}COO^{-}]

5.6 x
10^{-10} = x^{2} / (0.39 - x )

5.6 x
10^{-10} = x^{2} / 0.39 (Neglecting x term from
denominator since x << 0.39)

x^{2}
= 2.184 x 10^{-10}

x
= 1.48 x 10^{-5}

[OH^{-}] =
x = 1.48 x 10^{-5}
M

pOH = - log
[OH^{-}] = -log(1.48 x 10^{-5}) = 4.83

pH + pOH =14

pH = 14 - pOH = 14 - 4.83 = 9.17

a) Calculate the pH of a 0.22-M
acetic acid solution. Ka (acetic acid) = 1.8 × 10-5
pH = ____
b) You add 83 g of sodium acetate to 1.50 L of the 0.22-M acetic
acid solution. Calculate the new pH of the solution. (Ka
for acetic acid is 1.8 × 10-5).
pH = _____

Calculate the pH of a 0.59 M CH3COONa solution. (Ka for acetic
acid = 1.8 × 10^−5.)

Calculate the pH of a 0.42 M solution of sodium acetate,
CH3COONa. (Ka(acetic acid) = 1.8 x 10-5)

acetic acid (hc2h3o2) is a weak acid (ka= 1.8*10^-5). Calculate
the pH of a15.1 M HC2H3O2 solution.

a) If acetic acid is the only acid that vinegar contains
(Ka=1.8×10−5), calculate the concentration of
acetic acid in the vinegar. A particular sample of vinegar has a pH
of 2.95.
b) The acid-dissociation constant for benzoic acid (C6H5COOH) is
6.3×10−5. Calculate the equilibrium concentration of H3O+ in the
solution if the initial concentration of C6H5COOH is
6.2×10−2 M .
c) Calculate the equilibrium concentration of C6H5COO− in the
solution if the initial concentration of C6H5COOH is
6.2×10−2 M ....

What is the pH of a 0.010 M acetic acid solution? (Ka for acetic
acid is 1.76 x 10–5.)

Using the ionization constants for acetic acid (Ka= 1.8 x 10^-5)
and ammonia (Kb= 1.8 x10^-5), calculate the expected pH of the
above four solutions (0.100 M sodium acetate,
ammonium chloride, acetic acid and ammonia). Use the simplified
version instead of the quadratic equation.

The Ka of acetic acid (CH3CO2H) is 1.8 × 10–5. What is the pH of
a solution that is made up by dissolving 0.45 moles of sodium
acetate (CH3CO2Na) and 0.250 moles of acetic acid (CH3CO2H) in
enough water to make 1 L of solution?
1.80
4.74
5.00
6.12
7.33
4.48

You have a 250.-mL sample of 1.22 M acetic acid (Ka = 1.8 ×
10–5). Calculate the pH of the best buffer.

Calculate the pH of a solution of 0.25 M Acetic acid and 0.34M
Sodium acetate before and after 0.036 M NaOH is added to the
solution. Ka of HAc = 1.8 x 10 -5

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