Question

A 0.0202M solution of chloroacetic acid is 24.1% ionized at 25°C calculate the Ka for this acid

Answer #1

ClCH2COOH <------> ClCH2COO^{−} + H^{+}

The acid-dissociation equilibrium constant is:

Ka = [ClCH2COO^{−}][H^{+}]/[ClCH2COOH]

We can find [H^{+}] using the concentration and the percent
dissociation:

0.241 x 0.0202 = 4.86 x 10^{-3} M

To find [ClCH2COO^{-}], we use the 1:1 stoichiometry from
the above equation:

[ClCH2COO^{-}] = [H^{+}] = 4.86 x 10^{-3}
M

For the [HAc], we can subtract 4.86 x 10^{-3} M from the
given concentration of 0.02002M:

[HAc] = 0.02002 - 4.86 x 10^{-3}= 0.01533M

Now we can calculate Ka by inserting the concentrations
found,

Ka = [(4.86 x 10^{-3})( 4.86 x 10^{-3})]
/0.01533

Ka = 1.54 x 10^{-3}

Calculate the pH of 0.100 molar solution of sodium
chloroacetate. Ka for chloroacetic acid = 1.4 x 10^-3

A 0.200 M solution of a weak acid, HA, is 9.4% ionized. Using
this information, calculate Ka for HA.

Calculate the pH at 25° C of a 0.055 M solution of a weak acid
that has Ka = 1.4 × 10−5.

Calculate the pH at 25°C of a
0.075 M solution of a weak acid that
has Ka = 1.3
×10−5.

What is the percent dissociation of a 0.0217 M solution of
chloroacetic acid, (HC2H2ClO2 )? (Ka = 1.35 × 10−3 )

A 1.25-molar solution of a weak monoprotic acid is 9.2% ionized.
Calculate the pH of the solution.

The Ka of propanoic acid (C2H5COOH) is 1.34×10−5. Calculate the
pH of the solution and the concentrations of C2H5COOH and C2H5COO−
in a 0.331 M propanoic acid solution at equilibrium.
pH=
C2H5COOH]=
[C2H5COO−]=
At 25 °C,25 °C, how many dissociated H+H+ ions are there in 357
mL357 mL of an aqueous solution whose pH is 11.95?

The Ka of a citric acid is 1.95x10^-5 at 25 C. what is the pH of
a 0.35 M aqueous solution of citric acid?

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 = _____

The pH at 25 °C of an aqueous solution of the sodium salt of
hydrocyanic acid (NaCN) is 11.14. Calculate the concentration of
CN- in this solution, in moles per liter. Ka for HCN is equal to
6.2×10-10.

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