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3. (2 Points) Suppose 53.0 mL of 0.320 M KOH is mixed with 28.0 mL of 0.250 M HNO3. What is the concentration of OH− ions after the reaction goes to completion? Express your answer in mol/L.

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The number of moles of excess hydroxide ions at the end of the reaction is 0.123 mol/L.

The equation of the reaction is;

KOH(aq) + HNO3(aq) ------->KNO3(aq) + H2O(l)

Number of moles of KOH = 53.0/1000 L ×  0.320 M = 0.01696 moles

Number of moles of HNO3 = 28/1000 L× 0.250 M = 0.007 moles

We can see that the acid, HNO3 is the limiting reactant so 0.007 moles of acid reacts with 0.007 moles base.

Number of moles of excess base = 0.00996 moles

Total volume of solution = 53.0 mL + 28.0 mL = 81 mL or 0.0081 L

Number of moles of excess base = 0.00996 moles/0.0081 L = 0.123 mol/L

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