Answer to Question #235046 in General Chemistry for ruby

Question

concentration if 34.50 mL of the acid is required to neutralise 25.0 mL of a solution of 0.150 M sodium carbonate solution.

Answer 1

$Na_2CO_3+2HCL \to2NaCl+CO_2+H_2O$

Moles of sodium carbonate $=\frac{25.0×0.150}{1000}=0.00375moles$

Moles of acid $=2×0.00375=0.0075moles$

Molarity of acid $=\frac{0.0075}{0.03450}=0.217M$