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Chapter 4: Q4.54P (page 180)

Sodium hydroxide is used extensively in acid-base titrations because it is a strong, inexpensive base. A sodium hydroxide solution was standardized by titrating 25.00 mL of 0.1528 M standard hydrochloric acid. The initial buret reading of the sodium hydroxide was 2.24 mL, and the final reading was 39.21 mL. What was the molarity of the base solution?

Short Answer

Expert verified

The molarity of the base solution is 0.103 M NaOH.

Step by step solution

01

first write down the balanced molecular equation:

NaOH(aq)+2HCl(aq)鈫�NaCl(aq)+H2O(l)

02

To find out the molarity of the base solution, first, we have to calculate the moles of sodium hydroxide:

Given values are:

volumeofstandard0.1528MHClsolution=25mlinitialburetereading=2.24mlfinalburetereading=39.21ml

Moles ofSodium hydroxide:

molesofNaOH=(25mlH2SO4)(1lit1000ml)(0.1528molHCl1litHClsolution)(1molNaOH1molHCl)

=3.82x10-3molNaOH

The molarity of the base solution is:

molarityofNaOH=molesofNaOHvolumeofNaOH

=3.82x10-3molNaOH(39.21ml-2.24ml)(1lit1000ml)

=0.103MNaOH

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Most popular questions from this chapter

In a titration of HNO3, you add a few drops of phenolphthalein indicator to 50.00 mL of acid in a flask. You quickly add 20.00 mL of 0.0502 MNaOH but overshoot the end point, and the solution turns deep pink. Instead of starting over, you add 30.00 mL of the acid, and the solution turns colorless. Then, it takes 3.22 mL of the NaOH to reach the end point. (a) What is the concentration of the HNO3solution? (b) How many moles of NaOH were in excess after the first addition?

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