Question

1. A student is given a buffer that contains [HA] and [A- ]. The student is given the concentration of [HA] and need to solve for the concentration of [A- ]. Should they titrate with HCl or NaOH? Explain your reasoning

2. At an equivalence point of 35.0 mL, how many moles of [HA] are present in the buffer if you titrated with 0.125 M NaOH?

3. A student titrated 25.0 mL of a buffer with 0.126 M HCl. The observed data is below. Hand draw a graph of the data (on graph paper) and determine the concentration of the weak base in the buffer. Thoroughly document your logic. (Hint: this is not a dilution – this is a titration)

Vol of HCl Added (mL) pH Vol of HCl Added (mL) pH 0.00 4.45 15.00 3.58 1.00 4.40 16.00 3.48 2.00 4.35 17.00 3.36 3.00 4.30 18.00 3.25 4.00 4.25 19.00 3.10 5.00 4.20 20.00 2.80 6.00 4.15 21.00 2.50 7.00 4.10 22.00 2.30 8.00 4.05 23.00 2.20 9.00 4.00 24.00 2.15 10.00 3.94 25.00 2.10 11.00 3.88 26.00 2.00 12.00 3.81 27.00 1.99 13.00 3.75 28.00 1.98 14.00 3.66 29.00 1.96

Answer 1

I am going to answer part 1 and 2 because part 3 is impossible to read the data. So if you want you can post a new question with the part 3. Organizing the data in a table so we can read it and help you with that!

1. Normally when you talk about HA you are talking about an acid compound. To titrate an acid you need to neutralize it, to be able to reach the equivalence point. So to neutralize an acid you only can do it with a base, in this case NaOH, because if you add more acid you will only make the pH more acidic and it will never be neutralize.

2. at an equivalence point, the moles of the acid and the base are equal. So calculating the moles of NaOH in that volume we can obtain the moles of the HA reacting.

Then we have V= 0.035 L M= 0.125 mol/L

n= 0.035L * 0.125 mol/L

n= 0.00438 mol of NaOH = 0.00438 mol of HA.