Question

Calculate the pH of a 0.150 M solution of benzoic acid
(C_{6}H_{5}CO_{2}H) if

K_{a}= 6.3 x 10^{-5}for the
acid. C_{6}H_{5}CO_{2}H(aq) ⇌ H^{+}(aq)
+ C_{6}C_{5}CO_{2}^{-}(aq)

Answer #1

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5CO2-

0.15 0 0

0.15-x x x

Ka = [H+][C6H5CO2-]/[C6H5COOH]

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 ((6.3*10^-5)*0.15) = 3.074*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.3*10^-5 = x^2/(0.15-x)

9.45*10^-6 - 6.3*10^-5 *x = x^2

x^2 + 6.3*10^-5 *x-9.45*10^-6 = 0

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

a = 1

b = 6.3*10^-5

c = -9.45*10^-6

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 = 3.78*10^-5

roots are :

x = 3.043*10^-3 and x = -3.106*10^-3

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

x = 3.043*10^-3

So, [H+] = x = 3.043*10^-3 M

use:

pH = -log [H+]

= -log (3.043*10^-3)

= 2.5167

Answer: 2.52

Benzoic acid, C6H5CO2H, is a weak acid (Ka=6.3x10^-5). Calculate
the initial concentration (in M) of benzoic acid that is required
to produce an aqueous solution of benzoic acid that has a pH of
2.54

Question:
Consider a .045 M solution of benzoic acid
(C6H5CO2H) whose Ka
value is 6.5 x 10-5.
A). write the acid-base reaction equation for this acid with
water:
B). write the equilibrium expression (Ka) for this
reaction:
C). Calculate the concentration of benzoic acid, benzoate ion
(the conjugate base), and hydronium ion for this solution.
D). Calculate the pH of this solution.
E). Calculate the concentration of hydroxide ions in this
solution.

How many grams of benzoic acid, C6H5CO2H, are in 3.00 L of a
solution whose pH is 2.8? Ka=6.5*10-5. The answer is
14.7 g benzoic acid but I do not know how they got to that answer.
Can someone please show me how to do this question?

In a solution containing benzoic acid (Ka = 6.3*10^-5),
propanoic acid (Ka = 1.3*10^-5), formic acid (Ka = 1.8*10^-4), and
hydrazoic (Ka = 2.5*10^-5) acid buffered to pH 4.373, what percent
of hydrazoic acid is protonated?

Find the pH of a 0.380 M aqueous benzoic acid solution. For
benzoic acid, Ka=6.5⋅10−5.

Assume you dissolve 0.235 g of the weak acid benzoic acid,
C6H5CO2H, in enough water to make 9.00 ✕ 10^2 mL of solution and
then titrate the solution with 0.163 M NaOH. C6H5CO2H(aq) + OH-(aq)
C6H5CO2-(aq) + H2O(ℓ) What are the concentrations of the following
ions at the equivalence point? Na+, H3O+, OH- C6H5CO2- What is the
pH of the solution?

Calculate the pH during the titration of 25.0 ml of 0.150 M
benzoic acid with 0.150 M NaOH after the following additions of
titrant: 15.0 ml; 25.0 ml; 40.0 ml.

Assume you dissolve 0.235 g of the weak acid benzoic acid,
C6H5CO2H, in enough water to make 7.00 ✕ 102 mL of solution and
then titrate the solution with 0.153 M NaOH.
C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O(ℓ)
What are the concentrations of the following ions at the
equivalence point? Na+, H3O+, OH- C6H5CO2-
M Na+
M H3O+
M OH-
M C6H5CO2
- What is the pH of the solution?

A buffer solution is prepared by adding 0.5 moles of benzoic
acid (C6H5COOH, Ka = 6.3 x 10-5) and 0.5 moles of potassium
benzoate (C6H5COOK) to enough water to make a 1L solution. How many
grams of barium hydroxide is needed to shift the pH of this
solution to 6? (This question was previously answered. The answer
is not 19.4

Acid Ka
H2PO4- 6.29 x 10-8
NH4+ 5.68 x 10 -10
HCl >> 1
Calculate the pH of a 0.150 M solution of Na2HPO4.
Calculate the pH of a 0.150 M solution of NH4NO3.
Calculate the pH of a 0.150 M solution of KCl.

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