Question

To answer the following questions, refer to this titration curve. The titration was performed on a mixture of HCl and H3PO4 with 0.099M NaOH. You must show your work for full credit. What are the coordinates of the first and second equivalence points? How many moles of acid were neutralized at the first equivalence point? How many moles of acid were neutralized at the second equivalence point (not including the moles neutralized at the first equivalence point)? What is the composition of the acid mixture (ratio of HCl to H3PO4)?

Latest Acid-Base Titration Volume (mL) pH d1 2.36 2.36 239 246 2.57 2.65 3.4 4.5 5.5 as a o o o one wм- б 6.5 7 12 1.3 74 7.5 77 7.9 8.2 8.5 9 9.5 10 10.5 11 11.15 112 11.3 11.4 11.5 11.6 11.8 12 12.3 2.88 3.12 3.51 3.96 5.00 5.32 5.76 6.05 623 6.44 6.60 6.86 7.07 129 7.56 7.94 8.13 8.24 8.53 9.22 9.78 10.11 10.50 10.63 10.83 (pH/vol) 0.009 0.01g 0.042 0.072 0.113 0.175 0.247 0.392 0.657 1.344 4.126 6.028 3.910 2292 1.247 0.80g 0.620 0.522 0.47 0.451|| 0.505 0.642 0.851|| 1.424 2.857 4829 5.705 4.277 2.378 1.31 1|| 0.720 0.524, pH 15 Volume (mL)

Answer 1

The titration was performed on a mixture of HCl and H3PO4 with 0.099M NaOH.

What are the coordinates of the first and second equivalence points?   :

At equivalence point , there is sudden spike in pH :

• 1st equivalence point : pH : 5.00 ; Volume of 0.099M NaOH added : 7.3 mL
• 2nd equivalence points : pH : 9.22 ; Volume of 0.099M NaOH added : 11.4 mL

How many moles of acid were neutralized at the first equivalence point?

moles of Acid neutralized = moles of NaOH added

=   0.099 mol/L * (7.3 mL/1000 ml/L) = 7.23*10-4 moles

How many moles of acid were neutralized at the second equivalence point (not including the moles neutralized at the first equivalence point)? :

moles of acid were neutralized at the second equivalence point = moles of NaOH added

= 0.099 mol/L * ((11.4 mL - 7.3 ml) /1000 ml/L) = 4.06*10-4 moles

What is the composition of the acid mixture (ratio of HCl to H3PO4)? :

1st equivalence point = HCl + H3PO4

2nd equivalence point = H3PO4

thus moles of HCl = 7.23*10-4 moles - 4.06*10-4 moles= 3.17*10-4 moles

Moles of H3PO4 = 4.06*10-4 moles

Ratio of Acids in mixture = H3PO4 / HCl = (4.06*10-4 moles/3.17*10-4 moles)   = 1.28