Determining lon concentrations in mixtures (no precipitates): Find the [OH] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH)2.

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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How to calculate the [OH]
Determining lon concentrations in mixtures (no
precipitates):
Find the [OH-] when 50.0 mL of 0.20M NaOH is mixed
with 250.0 mL of 0.30M Sr(OH)2.
Both of these solutions are soluble (check solubility Table). There is NO
compound of Low Solubility involved here. So no equilibrium and no Ksp!
In this type of problem:
-use the Dilution Formula to find the final [NaOH] in the mixture
(final volume is 300.0 mL)
-dissociate the NaOH to find the [OH'] in NaOH
-use the Dilution Formula to find the final [Sr(OH),]
-dissociate the [Sr(OH),] to find the [OH] in [Sr(OH)2]
-add up the [OH] from NaOH and the [OH] from [Sr(OH),] to find the
final [OH] .
-THE ANSWER IS [OH] =
(0.53 M)
%3D
2021/7/5
UST DEPARTMENT OF CHEMICAL ENGINEERING
37
Transcribed Image Text:Determining lon concentrations in mixtures (no precipitates): Find the [OH-] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH)2. Both of these solutions are soluble (check solubility Table). There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH'] in NaOH -use the Dilution Formula to find the final [Sr(OH),] -dissociate the [Sr(OH),] to find the [OH] in [Sr(OH)2] -add up the [OH] from NaOH and the [OH] from [Sr(OH),] to find the final [OH] . -THE ANSWER IS [OH] = (0.53 M) %3D 2021/7/5 UST DEPARTMENT OF CHEMICAL ENGINEERING 37
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