# A sample of 1.00g of potassium carbonate is dissolved in 100mL water. A titration of this sample requires mL of nitric acid to reach the endpoint. What is the concentration of the Nitric Acid?

## The concentration or answer is 0.598 mol/L but I don't quite understand how that question was answered. If work could be shown it would help a lot. Thanks.

Jan 10, 2018

We assess the stoichiometric reaction....

${K}_{2} C {O}_{3} \left(a q\right) + 2 H N {O}_{3} \left(a q\right) \rightarrow 2 K N {O}_{3} \left(a q\right) + C {O}_{2} \left(g\right) \uparrow + {H}_{2} O \left(l\right)$

#### Explanation:

And so we work out the molar quantity with respect to carbonate....

${n}_{\text{potassium carbonate}} = \frac{1.00 \cdot g}{138.21 \cdot g \cdot m o {l}^{-} 1} = 7.24 \times {10}^{-} 3 \cdot m o l$ ..

And this represents HALF an equivalence of added nitric acid...

But you have not quoted the volume of nitric acid you used for the titration. And so, we are left in ignorance. We cannot assess $\left[H N {O}_{3}\right]$ without knowledge of the volume of titrant....

Edited to add....now we gots a volume of titrant $\equiv 24.20 \cdot m L$. Note that for standard burettes, I think 4 significant figures are absolutely appropriate....with a pair of spexs I can still read up to $0.01 \cdot m L$...certainly this is within the standard error of $0.05 \cdot m L$.

And so we set up the concentration quotient....

$\frac{2 \times 7.24 \times {10}^{-} 3 \cdot m o l}{24.20 \times {10}^{-} 3 \cdot L} \equiv 0.598 \cdot m o l \cdot {L}^{-} 1$

i.e. $\left[H N {O}_{3}\right] = 0.598 \cdot m o l \cdot {L}^{-} 1$....

Have you followed my approach? I don't think I have really nailed it. I work from the stoichiometric equation....and this allows me to assess the stoichiometric equivalence.