A sample of cola drink (exactly 100.0 mL) was titrated with 0.1025 N NaOH. If th

Question

A sample of cola drink (exactly 100.0 mL) was titrated with 0.1025 N NaOH. If the first equivalence point occurred after 13.11 mL of base were added, and the second equivalence point occurred after 28.55 mL of base, calculate the concentration of H3PO4 and H2PO4- in the cola sample (Hint: Where should the second equivalent point have occurred if only H3PO4 were present? Since the observed endpoint occurs beyond the point where we expect it to occur for a sample of pure H3PO4 in water, some H2PO4- must also be present along with the H3PO4 in the sample.)

Explanation / Answer

Relevant equations H3PO4 + 2OH --> 2H2O + HPO4

The first eq point is at 13.11 mL and that titrates the first H of the H3PO4. If the sample contained ONLY H3PO4, then the second eq point would be at 13.11 x 2 or 26.22 mL to titrate the second H. So the difference between 28.55 and 26.22 must be due to the H2PO4^

I subtracted the first eq Volume from the second: 28.55-.26.22 ml and got 2.33 L =2.33*10-3
I times this Volume by [NaOH] and got NaOH moles= 2.33*10-3 x 0.1029= 2.39757*10-4
Since 2 moles NaOH=1 mol biphosphate--> 2.397*10-4 x 1/2= 1.1985 x 10-4
I divide this by 0.1 L to get [H2PO4]=1.1985 x 10-3

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