PRELIMINARY QUESTIONS (To be answered BEFORE coming to lab.) 1. Given the reacti

Question

PRELIMINARY QUESTIONS (To be answered BEFORE coming to lab.) 1. Given the reaction M(H202 0H- a) For each ml of 0.1 M solution of the metal ion, how many ml of 1 M Na0H would M(H20)4 (0H)2 2 H20 be required for complete reaction? Show your reasoning. b) What volume constitutes an excess of 0H? 2. Show the calculation of the equilibrium [OH] in 1.0 M NH3 corresponding pH. Then calculate the 3. The pH of an ideal 0.10 M NaHCO3 solution is 8.31. In the list of bases on p. 77, HC03 is shown as being a better base than the 0Ac ion. a) Show the calculation of the pH of a 0.10 M Na0Ac solution b) Explain why the Na0Ac solution is more basic than the corresponding NaHCO3 solution. EXPERIMENTAL 2+ 2+ 3+ Use Nichrome wire as before to perform flame tests on solid samples of the nitrates listed in the table. Record your observations. How long did the color last?

Explanation / Answer

Part 1 a- For each mole of metal ion, 2 moles of NaOH is required as per the stoichiometry of the reaction.

1 ml 0.1 molar solution of metal ions will contain 0.1x 10-3 moles, hence 0.2x 10-3 moles of NaOH is required.

Since NaOH solution contains one mole NaOH per liter, 0.2 ml of NaOH solution would be required for complete reaction.

Part 2- For dissociation of NH3 ,

Pre equilibrium- NH3 (1 Mole) + H2O ---> NH4+ (0 moles)+ OH- (0 moles)

Equilibrium- NH3 (1 - X Mole) + H2O ---> NH4+ (X moles)+ OH- (X moles)

Kb for NH3 = [NH4+][OH-] / [NH3]

1.8x10-5 = [X][X] / [1]

1.8x10-5 = X2

X = 0.424 X 10-3 = number of equilibrium OH- ions in the solution

pOH- = - log [OH-]

= - (- 3.3726)

= 3.3726

pH+ pOH- = 14

hence, pH = 10.627

Part 3 a

pH of 0.1 M NaOAc

Kb for NaOAc is = [AcOH][OH-]/[NaOAc]

5.376 × 10-10 = [AcOH][OH-]/[NaOAc]

[OH-] = 7.332 × 10-6 M

pH = 8.865

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