We’re going to titrate formic acid with the strong base, NaOH. There is initiall

Question

We’re going to titrate formic acid with the strong base, NaOH. There is initially 100. mL of 0.50 M formic acid and the concentration of NaOH is 1.0 M.

5) What volume of NaOH is required to reach the equivalence point?                            ___________

6) What is the pH at the equivalence point?                                                                 ___________

7) What is the pOH at the equivalence point?                                                              ___________

8) If, instead of NaOH being added, 0.05 moles of HCl is added by                                   __________%

     bubbling the gas through the solution. Assume that the volume has

     not changed. What is the percent dissociation of formic acid?

Part 2. Examine the following compounds. Determine if they are acidic, basic, or neutral when dissolved in water.

A) N(CH3)H3Br                                                                                           A)____________

B) Fe(NO3)3                                                                                                                                             B)____________

C) NaNO3                                                                                                                                                  C)____________

D) K3PO4                                                                                                  D)____________

E) HClO2                                                                                                                                                    E)____________

Part 3. Identify the labeled points in the following titration curve.

A ___________________________

B ___________________________

C ___________________________

D ___________________________

E ___________________________

F ___________________________

G ___________________________

H ___________________________

I ___________________________

J ___________________________

Explanation / Answer

5) At equivalence mmoles of acid = mmoles of base

0.50Mx 100mL = 1.0 M xV mL

thus volume of NaOh = 50 mL

6) pH at equivalence At equivalence only HCOONa if formed

and [salt] = 50mmol/ 150mL= 1/3

and pH of solution = 1/2[pKw + pKa + log C]

= 1/2 [14 + 3.75 + log 1/3]

= 8.6364

7) pOh of solution = 14 - pH

= 14 - 8.6364

= 5.3635

8) HCOOH --------> HCOO- + H+

C 0 0 initial

0.5(1-x) 0.5x 0.05

Thus Ka = 1.8x10-4 = (0.5 x)0.05/ 0.5 (1-x)

thus x = 3.613 x10-3 M

part 2

a) N(CH3)H3Br methylammonium bromide - salt of weak base and strong acid - acidic in solution due to cationic hydrolysis

b) Fe(NO3)3 - salt of weak base and strong acid - acidic ues to cationic hydrolysis

c) NaNO3 salt of strong acid and strong base - neutral

d) K3PO4 - salt of strong base and weak acid - basic due to anionic hydrolysis

e) HClO2 - an acid

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