Question

A 27.8 mL sample of 0.254 M trimethylamine, (CH3)3N, is titrated with 0.275 M hydrobromic acid After adding 10.9 mL. of hydrobromic acid, the pH is Use the Tables link in the References for any equilibrium constants that are required The following graph shows the pH curve for the titration of 25 mL. of a 0.1 M monoprotic acid solution with a 0.1 M solution of a monoprotic base. 14 pH 12 10 0 5 10 15 20 25 30 35 40 45 50 mL. of 0.1 M base added (1) The plł curve represents the titration of aacid with abase (2) Choose a suitable indicator for the endpoint of the titration from the following pulldown list Malachine green Medryl erange Brsocresol green The equilibrium concentration of lead ion in a saturated lead chromate solution is Write the Ksp expression for the sparingly soluble compound zinc phosphate, Zny(PO)2 If either the numerator or denominator is 1, please enter 1

Answer 1

Question a: titration of trimethylamine with hydrobromic acid

In this titration, the reaction is

H30 . (aq) (aq)

Data:

	Concentration	Volume	Amount
(CH3)3N	0.254 M	27.8 mL	7.061 mmol
HBr	0.275 M	10.9 mL	2.998 mmol

The amount of (CH₃)₃N and HBr is calculated with the definition of Molarity:

moles moles Molarity x volume in L Molarity - volume in L

moles (CH3)3N 0.254-x 0.0278 L 0.00706 1mol 7.061 mmol

mol moles HBr= 0.275_-x 0.0 109 L 0.00299&nol = 2.998 mmol

The initial amount of H₃O⁺ that reacts with the trimethylamine is obtained from the hydrobromic acid (reaction below) because 1 mol of HBr produces 1 mol of H₃O⁺. Then 2.998 mmol of HBr produces 2.998 mmol of H₃O⁺.

r13 (a

Then, according to reaction summary,

		(CH3)3Na«a		3 (aq		33 (aq	+	H20)
Initial		7.061 mmol		2.998 mmol		0 M		--
Change		- 2.998 mmol		- 2.998 mmol		+ 2.998 mmol		--
Final		4.063 mmol		0 mmol		2.998 mmol		--

So, after addition of 10.9 mL of HBr, there is 4.063 mmol of (CH₃)₃N and 2.998 mmol of (CH₃)₃N⁺ in the solution. The concentration of each one is:

moles Molarity - volume in L

4.063 mmol [(CH3)3N] -0.105 AM 27.8mL+10.9 mL

2.998 mmol0.077 M [(CH3)3NH+] = 27.8 mL + 10.9 mL

These two compound are conjugated pair Acid-Base, according to reaction:

with an equilibrium constant,

[(CH3)3NH [OH] [(CH3 )3N]-_ = 6.4x10-5

The presence of (CH₃)₃N and (CH₃)₃N⁺ indicates that the solution acts like a basic buffer, and the pOH could be calculated with the Hendersson-Hasselbach equation:

[(CH3)3NH-] [(CH3)3N]

0.077 M 0.105 M/-4.06 DOH _ log(6.4x10-*) + log

Then the pH is

pH 14-рон 14-4.06 = 9.94

Conclusion: after adding 10.9 mL of hydrobromic acid, the pH is 9.94

Question b: titration curve

1. This pH curve represents the titration of a weak acid with a strong base

The x-axis of the curve indicating the addition of a base and the small interval of pH at the equivalence point indicates that the sample is a weak acid. And for the titrations of weak acid, strong bases are used.

2. The best indicator for the endpoint of this titration is thymol blue

The best indicator has a transition interval of color near to the pH at the equivalence point. In this case, the equivalence point is at pH 7, and the only indicator with significative change of color at this pH is thymol blue

Question c: lead chromate solution

A saturated solution of lead chromate is expressed by this reaction:

4 (aq)

with an equilibrium constant,

K,,-[Pb +2][Cr04-2] = 2.8x10-13

According to reaction summary,

		PbCr04(s)	+	H20)		Pb+2(aa)	+	CrO4 (aa) -2
Initial		--		--		0 M		0 M
Change		--		--		+ x		+ x
Final		--		--		x M		x M

Then,

K,,-[Pb +2] [cro,--] = (x) x (x)-x-= 2.8x10-13

2.8x10-13-5.29:10-7 M x =

The concentration of lead ion in a saturated lead chromate solution is 5.29x10^-7 M

Question d: zinc phosphate

The solubility of zinc phosphate is expressed by the reaction:

Znz(P04)23 Zn+2(a + 2 PO4 (сајт ZP04

And the equilibrium constant is

Zn+2]3 [p-312 sp