8 and 10 Please! 8. (10) You have discovered a new compound that you think may b

Question

8 and 10 Please!

8. (10) You have discovered a new compound that you think may be capable of undergoing "Cold Fusion." But you don't know the molecular weight of this compound, and you need to determine it before you can go win the Nobel prize and collect billions from patents. So, you pack an anion exchange column and charge it with hydroxide OH). You suspect this compound in water and pass it over the column, which exchanges 2 OH for every 1 B2 You then titrate your compound has the formula AB A3, B2). You dissolve 0.030 g of the resulting solution with 0.1 MHCL The initial reading is 0,5 ml and the final reading is 22.1 mL. What is the molecular weight of your compound? . 03 ZZ.1-.521.6 ny mot I L ) Fred and George Weasely (from the Harry Potter series) own a joke shop. One of their new holiday is bottled water that smells like dead fish. To do this, they added 15 m (0.015 g) of trimethylamine CHs)b) into IL of water. The Ks of trimethylamine is 64 x 10%. What is the resulting pH of the solution? I C

Explanation / Answer

[8] HCl volume used in titration = 22.1 - 0.5 = 21.6 ml

Molarity of HCl solution = 0.1 M

Number of millimoles of HCl used = 0.1 M * 21.6 ml = 2.16 mmol

Since, acid-neutralization reaction is H+ + OH- ----> H2O

therefore, Millimoles of OH- present in solution = 2.16 mmol

As, 1 molecule of B2- was exchanged for 2 OH-. It means 2 mmol of OH- was released for 1 mmol of B2-. Therefore, 2.16 mmol of OH- ion must have been released by 2.16/2 = 1.08 mmol of B2- ion.

Now, A2B3 ----> 2A3+ + 3B2-

it means 3 moles of B2- is produced by 1 mol of A2B3.

therefore, 1.08 mmol of B2- was released by (1 mol/3 mol)*1.08 mmol = 0.36 mmol of A2B3

Since, 0.36 mmol of A2B3 in solution is from 0.030 g of A2B3 used to make solution.

so, 0.36 mmol of A2B3 weight is 0.030 g

therefore, 1 mol of A2B3 weight = (0.030 g /0.36*10-3) = 83.33 g

Hence, molecular weight of A2B3 is 83.33 g/mol

[10] weight of trimethylamine used = 0.015 g

Molecular weight of trimethylamine = 59.11 g/mol

so, number of moles of trimethylamine used in 1.0 L water = 0.015 g / 59.11 g mol-1 = 2.54 * 10-4 mol

Thus, molarity of trimethylamine = 2.54 * 10-4 mol / 1.0 L = 2.54 * 10-4 M

Given, trimethylamine Kb = 6.4 *10-5

As, trimethylamine is a weak base, [OH-] = (Kb Cb)1/2

using Values of Kb and concentration of trimethylamine (Cb), we have

[OH-] = (Kb Cb)1/2 = (6.4 *10-5 * 2.54 * 10-4 )1/2 = (1.6256 * 10-8)1/2 = 1.275 *10-4 M

so, pOH = -log[OH] = -log(1.275 *10-4) = -(-3.894) = 3.894

Thus, pH = 14 - pOH = 14 - 3.894 = 10.106

Hence, pH of solution is 10.106

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