Problem: A solution of 116 mL of 0.180 M KOH is mixed with a solution of 260 mL of 0.210 M NiSO4.What is the concentration of SO42- that remains in solution?

FREE Expert Solution

Step 1

2 KOH + NiSO4 → Ni(OH)2 + K2SO4

116 mL ×1×10-3 L1 mL ×0.180 mol KOH1 L×1 mol K2SO42 mol KOH = 0.01044 mol K2SO4


260 mL ×1×10-3 L1 mL ×0.210 mol NiSO41 L×1 mol K2SO41 mol NiSO4 = 0.0546 mol K2SO4

KOH → limiting reactant


82% (260 ratings)
View Complete Written Solution
Problem Details

A solution of 116 mL of 0.180 M KOH is mixed with a solution of 260 mL of 0.210 M NiSO4.

What is the concentration of SO42- that remains in solution?

Learn this topic by watchingSolution Stoichiometry Concept Videos
All Chemistry Practice Problems Solution Stoichiometry Practice Problems
Q.A sample of 5.55 g of Mg(OH)2 is added to 25.2 mL of 0.180 M HNO3.Which is the limiting reactant in the reaction?
Q.Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant sol...
Q.A solution of 116 mL of 0.180 M KOH is mixed with a solution of 260 mL of 0.210 M NiSO4.What is the concentration of Ni2+ that remains in solution?
Q.A solution of 116 mL of 0.180 M KOH is mixed with a solution of 260 mL of 0.210 M NiSO4.What is the concentration of K+ that remains in solution?
See all problems in Solution Stoichiometry

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Solution Stoichiometry concept. You can view video lessons to learn Solution Stoichiometry. Or if you need more Solution Stoichiometry practice, you can also practice Solution Stoichiometry practice problems.