312 mL of 1.50 x 10^-2 M NaOH is used to titrate 225 mL of a monoprotic acid with unknown...

Question:

312 mL of {eq}1.50 \times 10^{-2} {/eq} M NaOH is used to titrate 225 mL of a monoprotic acid with unknown concentration to reach the equivalence point. What is the molarity of the original acid?

Equivalence Point:

While performing the acid-base titration reaction, acid is slowly added to the basic solution. We reach a point when the acid-base indicator changes its color. This indicates a point when the solution gets completely neutral. This state in acid-base titration is termed as the equivalence point of the neutralization reaction.

Answer and Explanation: 1

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Given:

  • The concentration and the volume of NaOH used in titration is {eq}M_1 = 1.5 \times 10^{-2} \ M \ and \ V_1 = 312 \ mL =0.312 \ L {/eq}
  • The...

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Equivalence Point: Definition & Calculation

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Chapter 12 / Lesson 11
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Learn about equivalence point and end point in titrations. Examine how to find equivalence points in a variety of ways, and discover the steps to do so.


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