Only need help with the back of the worksheet, post the first page as reference.

Question

Only need help with the back of the worksheet, post the first page as reference.

thank you will rate, please explain.

Briefly state how the graph indicates whether the acid is strong or weak and why a. ea 4. Consider the reaction at the half-equivalence point. L sample? b. How many moles of base are added at the half-equivalence point? c. Show the ICF table below for the reaction d. Calculate the pH of the solution at the half-equivalence point. Briefly state how the graph would give you the value of pKa without doing any calculations e. 5. Calculate the pH of the solution after the addition of 15.00 mL of the NaOH. a. Show the ICF table below. b. Show your pH calculation. c. Does your calculation match the graph? If not, justify your answer

Explanation / Answer

I will answer question 4 there (for now), and give you some guidance to do the question 5.

Assuming we have the same concentration of acid and base. we have:
[HA] = 0.0833 M; [OH] = 0.1 M

at 25 mL of acid, the moles:
moles = 0.0833 * 0.025 = 0.00208 moles
At the half equivalence point, we have:
moles base = 0.1 * 0.005 = 0.0005 moles added.

moles of acid in the half equivalence point: 0.0833 * 0.012 =

The ICE table:0.00099 moles
HA + OH- ----------> A- + OH-
i: 9.9x10-4   5x10-4  0 0
c: -5x10-4   -5x10-4  +5x10-4
e: 4.99x10-4 0 5x10-4

the expression to use will be:
pH = pKa + logA-/HA
pH = pKa + log(5x10-4/4.99x10-4)
pH = pKa + log1
pH = pKa

So, according to this, the pH = pKa and the pH according to the graph is 5. So this is the pH.

The graph will give us the value of pKa at the half equivalence point, because at this point only half of the moles of acid are used to react with the whole moles of base. At the end it will produce the half, and a 1:1 ratio of this, will give log1 = 0 so pH = pKa. Looking only the graph at the half equivalence point and see the pH we can know the pKa.

Now at 15 mL we are beyond equivalence point and the species in there, are excess of base, so the reaction taking place is:

A- + H2O <--------> HA + OH-   Kb

So plot an ice table there and solve for pH using the HH equation:
pH = pKa + logA/HA or pOH = pKb + logBH/B

Hope this helps

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