3. Ca(OH)2(s)⇄Ca2+(aq)+2OH−(aq) Ksp=5.5×10−6 at 298K The equilibrium in a saturated solution of Ca(OH)2 is represented above. In an experiment, a student places 5.0 g of Ca(OH)2(s) into 100.0 mL of distilled water and stirs the mixture. How would the results be affected if the student repeats the experiment but this time places 5.0 g of Ca(OH)2(s) into 100.0 mL of 0.0010 M NaOH(aq) instead of distilled water? a)Less solid will dissolve, because the larger value of [OH−] will cause the equilibrium position to lie farther to the right. b)Less solid will dissolve, because the larger value of [OH−] will cause the equilibrium position to lie farther to the left. c)More solid will dissolve, because the larger value of [OH−] will cause the equilibrium position to lie farther to the right. d)More solid will dissolve, because the smaller value of [OH−] will cause the equilibrium position to lie farther to the left.
3. Ca(OH)2(s)⇄Ca2+(aq)+2OH−(aq)
Ksp=5.5×10−6 at 298K
The equilibrium in a saturated solution of Ca(OH)2 is represented above. In an experiment, a student places 5.0 g of Ca(OH)2(s) into 100.0 mL of distilled water and stirs the mixture. How would the results be affected if the student repeats the experiment but this time places 5.0 g of Ca(OH)2(s) into 100.0 mL of 0.0010 M NaOH(aq) instead of distilled water?
a)Less solid will dissolve, because the larger value of [OH−] will cause the equilibrium position to lie farther to the right.
b)Less solid will dissolve, because the larger value of [OH−] will cause the equilibrium position to lie farther to the left.
c)More solid will dissolve, because the larger value of [OH−] will cause the equilibrium position to lie farther to the right.
d)More solid will dissolve, because the smaller value of [OH−] will cause the equilibrium position to lie farther to the left.
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