1. What volume in mililiters of a 0.111 M NaOH solution is required to reach the

Question

1. What volume in mililiters of a 0.111 M NaOH solution is required to reach the equivalance point in the complete titration of a 9.00 ml sample of 0.122 M H2SO4?

2. A solution has [H3O+] = 7.6 x 10 -5 M. Use the ion product constant of water (Kw=[H3O+][OH-]) to find te [OH-] of the solution.

3. A solution has [OH-]=2.1x10-2 M. Use the ion product constant of water to find the [H3O+] of the solution.

4. Classify each solution as acidic, basic or neutral accordong to its pH value.

a)pH= 8.0

b) pH= 7.0

c)pH= 3.5

d)pH=6.1

5. What is the pH of a solution with [H3O+]= 2.8 X 10-5 M?

a)10.48

b)1.00

c)-4.55

d)4.55

6. What is [OH-] in a solution with a pH of 9.55?

a) 2.82x10 4 M

b)4.45 M

c)3.55 x 10 -5 M

d)2.82 x 10 -10 M

Explanation / Answer

1. What volume in mililiters of a 0.111 M NaOH solution is required to reach the equivalance point in the complete titration of a 9.00 ml sample of 0.122 M H2SO4?

NaOH + H2SO4 <-> H2O + Na2SO4

balance

2NaOH + H2SO4 <-> 2H2O + Na2SO4

then

ratio is 1:2

mmol of acid = 0.122*9 = 1.098 mmol of H2SO4

then we need double amount of base

mmol obase = 1.098*2 = 2.196 mmol of base

V = mmol/M= 2.196/0.111 = 19.7837 mL

NOTE Consider posting multiple questions in mutliple set of Q&A.

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