Question

Enter your answer in the provided box. Sodium hydroxide is used extensively in acid-base titrations because it is a strong, inexpensive base. A sodium hydroxide solution was standardized by titrating 28.58 mL of 0.1851 M standard hydrochloric acid. The initial buret reading of the sodium hydroxide was 1.98 mL, and the final reading was 30.89 mL. What was the molarity of the base solution?

Answer 1

Answer:

Molarity of the base solution was 0.1830 M

Explanation:

Total volume of NaOH added for complete standardization = (final reading-initial reading) = (30.89-1.98) mL = 28.91 mL

Neutralization reaction between NaOH and HCl :

                                 $NaOH+HCl\rightarrow NaCl+H_{2}O$

So, 1 mol of HCl neutralizes 1 mol of NaOH

Moles of HCl added = $\frac{28.58\times 0.1851}{1000}$ moles

If molarity of NaOH was C (M) then moles of NaOH added is $\frac{C\times 28.91}{1000}$

$\frac{28.58\times 0.1851}{1000}$ = $\frac{C\times 28.91}{1000}$            

or, C = 0.1830

So, molarity of NaOH was 0.1830 M