An 18.4 ml solution of 0.350 M nitric acid reacts with 25.3 ml of a sodium carbonate solution according to the equation. How many moles of nitric acid and moles of sodium carbonate reacted? What is the molarity of the sodium carbonate solution?

1 Answer
Feb 6, 2018

We use the relationship #"concentration"="moles of solute"/"volume of solution"#

#[Na_2CO_3(aq)]=0.127*mol*L^-1#

Explanation:

And, as always, we write a stoichiometric equation to represent the reaction....

#2HNO_3(aq) +Na_2CO_3(aq) rarr 2NaNO_3(aq) + H_2O(l)+CO_2(g)uarr#

And so.......

#"moles of nitric acid"=18.4*mLxx10^-3*L*mL^-1xx0.350*mol*L^-1=6.44xx10^-3*mol.#

And given the stoichiometry, there were HALF this molar quantity with respect to #Na_2CO_3(aq)#... Agreed?

And so #[Na_2CO_3(aq)]=(1/2xx6.44xx10^-3*mol)/(25.3*mLxx10^-3*L*mL^-1)#

#=0.127*mol*L^-1#