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Stoichiometry and Chemical Solutions Study Guide (page 2)

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Updated on Sep 25, 2011

Example:

What is the final molarity if 75 mL of 12 M concentrated HCl was diluted to 500 mL?

Solution:

So, M1 = 12 M; V1 = 75 mL; M2 = ?; V2 = 500 mL

How much 3.0 M NaOH is needed to make 350 mL, 0.10 M NaOH solution?

So, M1 = 3.0 M; V1 = ?; M2 = 0.10 M; V2 = 350 mL

Part Per Million and Part Per Billion

A common gas and liquid concentration is parts per million and parts per billion. The "parts" measure the molecules per million or billion:

   1 ppm = 1 molecule per every 1,000,000 molecules

1 ppb = 1 molecule per every 1,000,000,000 molecules

These concentration units are typically used when the concentrations are extremely small and it is impractical to use percent or molarity.

Titration

A titration involves the measurement of a titrant with a known concentration when reacted with an analyte (the substance being analyzed). The equivalence point or stoichiometric point of the reaction is where enough titrant has been added to exactly react with the analyte. The equivalence point of any base is the amount in grams that can be neutralized by 1 mole of H+ ions. The equivalence point of any acid is the amount in grams that can be neutralized by 1 mole of OH ions. Usually, a substance called an indictor is added at the beginning to show a change by color in the equivalence point.

The equation M1V1 = M2V2 can be used to calculate the molarity of the analyte.

Example:

Exactly 45.23 mL of 19.2 M NaOH was titrated to the equivalence point with 75.0 mL of an unknown concentration of HCl. Calculate the molarity of the HCl solution.

Solution:

 

Practice problems for these concepts can be found at -  Stoichiometry and Chemical Solutions Practice Questions

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