Question

# 6. Consider the following balanced acid/base reaction: 2 CH3COOH +   Ba(OH)2 -> (CH3COO)2Ba   + 2 H2O...

6. Consider the following balanced acid/base reaction:

2 CH3COOH +   Ba(OH)2 -> (CH3COO)2Ba   + 2 H2O

A student pipetted 5.00 mL of a 0.224M solution of CH3COOH into an Erlenmeyer flask. The student calculated that 36.44 mL Ba(OH)2 was needed to neutralize the acid. Calculate the molarity of Ba(OH)2 in the unknown solution.

7. In an aqueous solution the [H3O+] is 2.77x 10-9.M.

a. Is this solution acidic, basic or neutral?

b. What is the [OH-]?

8. In an aqueous solution the [OH-] is 8.49 x 10-5 M

a. Is this solution acidic, basic or neutral?

b. Calculate the pH of this solution.

9. The measured pH of a solution is 3.85.

a. Is this solution acidic, basic or neutral?

b. Calculate the [H3O+]

c. Calculate the [OH-]

Q6) From the equation

2Ch3COOH + Ba(OH)2 -----------> (CH3COO)2Ba + 2H2O

V1M1/n1 = V2M2/n2 at equivalence where N1 and n2 are stoichomtric coefficients

Thus

5mL x 0.224M /2 = 36.44mL x M of Ba(OH)2

thus molarity of Ba(OH)2 = 0.01537 M

Q7)

As [H3O+] = 2.77x10-9 M which is less than [H3O+] in water (1.0x10-7) the solution is basic.

We know that ionic produc t of water = [H3O+] [OH-] = 1.0x10-14

Thus [OH-] = Kw /[H3O+]

= 1.0x10-14  /2.77x10-9 M

= 3.61x10-6 M

Q8) Given [OH-] = 8.49 x 10-5

thus the solution is basic as [Oh-] > 1.0x10-7

the pOH of solution = -log [Oh-]

= 4.07

and pH of the solution = 14 -pOh

= 9.93

Q9)

The pH of solution = 3.85

So it is acidic solution [solution with pH < 7 are acidic]

the [H+] in solution = antilog -3.85

= 1.41x10-4 M

and [OH-] = 1.0x10-14 / 1.41x10-4 M

= 7.09x10-11 M

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