Question

10.00mL of 1.000M HA (a weak acid) was mixed with 5.00mL of 0.500M NaOH. Then the mixture was diluted to 50.0.mL with distilled water. The pH of the solution read 5.00.

(PL4) Write a balanced equation for dissociation of HA(aq) in water.

(PL5) Write an expression for K for the equation above.

(PL6) Draw the appropriate ICE table leading to Ka. Fill in all the information. Hint : Calculate [H3O+] at equilibrium from pH.

(PL7) Calculate Ka.

(PL8) What is the difference between K and Ka? Is the K from question above Ka? Explain.

(PL9) Draw a data table you will be filling in during the lab. Include all the measurements that will be made in this lab and qualitative observation if needed.

Answer 1

ans)

4.

from above data that

the balanced equation is for dissociation of HA

HA(aq) <---> H+(aq) + A-(aq)

5.  

k = [H+][A-]/[HA]

6.

pH = -log[H+]

[H+] = 10^(-Ph) = 10^(-5)

= 1*10^(-5) M

total volume = 10+5+50 = 65 ml.

No of moles of HA = 10/1000*1 = 0.01 mole

molarity of HA = 10/65*1 = 0.154 M

               HA(aq) <--->    H+(aq)   +    A-(aq)

Initial       0.154            0            0

equilibrium 0.154 - 1*10^(-5) 1*10^(-5)   1*10^(-5)

k = (1*10^(-5)*1*10^(-5)) / (0.154 - 1*10^(-5))

    = 6.5*10^-10

8)

K = ka no difference.