Determine the pH of the resulting solution when 45.00mL of 0.1500M HClO2 is mixed with 35.75mL of 0.3575M NaOH.

Chemistry: An Atoms First Approach
2nd Edition
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Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter6: Types Of Chemical Reactions And Solution Stoichiometry
Section: Chapter Questions
Problem 126CP
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Determine the pH of the resulting solution when 45.00mL of 0.1500M HClO2 is mixed with 35.75mL of 0.3575M NaOH.

 

(i attatched the steps needed to solve this problem.)

1. Write out the double replacement reaction between the given acid and base listing the
products and balancing the equation.
2. Determine the type of acid base reaction. There are 3 possible types of reactions.
- Strong Acid + Strong Base
- Weak Acid + Strong Base
-Strong Acid + Weak Base
3. Determine which reactant, the acid or the base is in excess and which reactant is the
limiting reagent. This is done by taking the given molarity for the reagent and multiplying
by the given volume in liters. MxV = moles The reagent with the greatest number of
moles is the excess reagent and the reagent with the least number of moles is the limiting
reagent.
4. Determine how much in excess the excess reagent is by subtracting the moles of the
limiting reagent from the moles of the excess reagent.
Excess moles of excess reagent = (moles of excess reagent from step 3) - (moles of
limiting reagent)
5. Determine the total volume of the solution when the acid and base are mixed. This is
done by summing the volume of the acid and the volume of the base.
Vatah = Vosid + Vboss Be sure to convert this to liters.
basa
6. Determine the new molarity of the excess reagent by taking the excess moles of the
excess reagent (from step 4) and dividing by the total volume (from step 5) in liters.
Transcribed Image Text:1. Write out the double replacement reaction between the given acid and base listing the products and balancing the equation. 2. Determine the type of acid base reaction. There are 3 possible types of reactions. - Strong Acid + Strong Base - Weak Acid + Strong Base -Strong Acid + Weak Base 3. Determine which reactant, the acid or the base is in excess and which reactant is the limiting reagent. This is done by taking the given molarity for the reagent and multiplying by the given volume in liters. MxV = moles The reagent with the greatest number of moles is the excess reagent and the reagent with the least number of moles is the limiting reagent. 4. Determine how much in excess the excess reagent is by subtracting the moles of the limiting reagent from the moles of the excess reagent. Excess moles of excess reagent = (moles of excess reagent from step 3) - (moles of limiting reagent) 5. Determine the total volume of the solution when the acid and base are mixed. This is done by summing the volume of the acid and the volume of the base. Vatah = Vosid + Vboss Be sure to convert this to liters. basa 6. Determine the new molarity of the excess reagent by taking the excess moles of the excess reagent (from step 4) and dividing by the total volume (from step 5) in liters.
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