A standard nitric acid solution is prepared using 0.425 g of sodium carbonate, #Na_2CO_3#. What is the molarity of the acid if 33.25 mL are required to reach a permanent endpoint?

Question

7658 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-23T16:24:24

#sf(0.241color(white)(x)"mol/l")#

Start with the equation:

#sf(Na_2CO_(3(s))+2HNO_(3(aq))rarr2NaNO_(3(aq))+CO_(2(g))+H_2O_((l)))#

Find the no. moles of #sf(Na_2CO_3)#:

#sf(n_(Na_2CO_3)=m/M_r=0.425/105.99=0.004)#

From the equation you can see that the no. moles of #sf(HNO_3)# must be twice this amount.

#:.##sf(n_(HNO_3)=0.004xx2=0.008)#

#sf(c=n/v=0.008/(33.75/1000)=0.241color(white)(x)"mol/l")#