MUST SHOW ALL WORK TO RECEIVE POINTS 1) Concentrated H3PO4 is 85% H3PO4 by weigh

Question

MUST SHOW ALL WORK TO RECEIVE POINTS

1) Concentrated H3PO4 is 85% H3PO4 by weight; its density is 1.689 g/ml. How many ml of concentrated H3PO4 should be diluted to one liter to make a 2.5 molar solution

2) In the dertemination of Chromium in an ore, all the chromium is converted to Cr2O7^-2 dichromate ion. This is titrated with a standard Fe^2+ solution:

6Fe+2 + Cr2O7-2 + 14H+ yields 6Fe+3 + 2Cr3+ +7H2O

How many grames of chromium are present in an ore sample which required 31.4 ml of .0982 M Fe+2 solution to titrate it?

4) Calculate the final concentration of H+ or Oh- if 24.o ml of 1.95 M Hc is mixed with 34.0 ml of 2.30 M NaOH. Assume volumes are additive and that HCl and NaOh are completely ionized in solution.

HCl yields H+ + Cl-

NaOh yields Na+ + OH-

Explanation / Answer

1) 2.5 Molar solution , V= 1 liter , we have M = moles/Vol

hence moles = MV = 2.5 x1 = 2.5 moles

H3PO4 = 2.5 moles = 2.5 x Molar mass gm = 2.5x98 = 245 gm

assuming 85 % solution H3PO4 mass required = ( 100/85) x245 = 288.2353

H3PO4 vol = mass/density = 288.253/1.689 = 170.65 ml = 170.65 ml

hence 170.65 ml of H3PO4 is taken and diluted to 1liter mark by water to get 2.5 Molar solution

2) moles of Fe2+ = MV = 0.0982 x (31.4/1000) = 0.0030835

as per stochiometry in equation dichromate ions ( Cr2O72-) ions titrated are = 0.0030835/6 =0.000514

we know 2 Cr is required to convert to Cr2O72-

hence Cr moles = 0.000514x2 = 0.001028

Cr mass = 0.001028 x 51.996 = 0.05344 gm

3) at equivalence point moles of NaOH = moles of HZ

hence 0.1 x (37.1/1000) = 0.0037 = moles of HZ = 0.3125/Molar mass

Molar mass = 84.56 gm/mol

4) moles of H+ = 1.95 x ( 24/1000) = 0.0468

moles of NaOH - 2.3 x ( 34/1000) = 0.0782

net OH- moles = 0.0782-0.0468 = 0.0314

total vol = 24+34 = 58 ml = 0.058 liters

[OH-] = ( 0.0314/0.058) = 0.5414 M

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