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  3. 10.00ml of 1.000m ha (a weak acid) was mixed with

10.00mL of 1.000M HA (a weak acid) was mixed with 5.00mL of 0.500M NaOH. Then the...

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Chemistry

10.00mL of 1.000M HA (a weak acid) was mixed with 5.00mL of0.500M NaOH. Then the mixture was diluted to 50.0.mL with distilledwater. The pH of the solution read 5.00.

1) What is the chemical equation between HA and NaOH.

2) Calculate moles of each. What is the limiting reactant?

3) Calculate the moles of each left at the end of this reaction.Calculate the Molarities for each. (What is the total volume?)

4) Write a balanced equation for dissociation of HA(aq) inwater.

5) Write an expression for K for the equation above.

6) Draw the appropriate ICE table. Fill in all theinformation.

Hint : Calculate [H3O+] at equilibrium from pH.

7) Calculate K.

8) What is the difference between K and Ka? Is the K fromquestion above Ka? Explain.

Answer & Explanation Solved by verified expert
3.6 Ratings (661 Votes)

1. HA(aq) + NaOH(aq) < ----> NaA(aq) + H2O(l)

2. limiting reagent is NaOH.

3.
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

No of moles of NaOH= 5/1000*0.5 = 0.0025 mole.

molarity of NaOH = 5/65*0.5 = 0.0385 M

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

5.   k = [H+][A-]/[HA]

6. pH = -log[H+]

[H+] = 10^(-Ph) = 10^(-5) = 1*10^(-5) 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.
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