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# Chemistry: Equilibrium Concentration Sample Questions

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a) What is the expression for the eqilibrium constant K in terms of the concentrations of the products and reactants for the reaction between ammonia, (NH3 (aq)) (SMALL 3), WHICH REACTS WITH WATER TO PRODUCE AMMONIUM (NH4 + (aq))
and hydroxide (OH - (aq)) ions.

NH3 (aq) + H20 (l) = NH4+ (aq) + OH- (aq)

the enthalpy change is +3.5kj mol

Using the equilbrium expression (from above) calculate What the OH- (aq) ion concentration in an ammonia solution with an equilibrium concentration of
1.00 x 10-2 mol dm-3 (ie 10 to the power -2)
the concentration of water at 25 degrees celcius is 55.4 mol dm-3
and the value of the equilibrium constant K for the reaction is 3.19 x 10-7
(ie 10 to the power -7). the answer needs to be to 3 significant figures.

b) use the calculated value for OH- (aq) ion concentration together with the expression for the ion product of water to calculate the H+ (aq) ion concentration. Calculate the PH for the ammonia solution
Assume that Kw = 1.00 x 10-14 mol2 dm-6 (ie 10 to the power -14)
at 25 degrees celcius, the answer needs to be to 3 significant figures.

Thanks! I just don't seem to get this!

https://brainmass.com/chemistry/energetics-and-thermodynamics/chemistry-equilibrium-concentration-sample-questions-321405

#### Solution Preview

Thank you.

finding equilibrium constant, OH- concentration ................
1A)
What is the expression for the eqilibrium constant K in terms of the concentrations of the products and reactants for the reaction between ammonia, (NH3 (aq)) (SMALL 3), WHICH REACTS WITH WATER TO PRODUCE AMMONIUM (NH4 + (aq))
and hydroxide (OH - (aq)) ions.

NH3 (aq) + H20 (l) = NH4+ (aq) + OH- (aq)

the enthalpy change is +3.5kj mol

using the equilbrium expression (from above) ...

#### Solution Summary

This solution provides assistance with the problem regarding equilibrium concentration.

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