# Calculating concentration of H+ Ions using Ka Value [closed]

I have this homework question that I am a bit stuck on for an answer. If anyone could help that would be great.

Question:

For a $\pu{0.2M}$ $\ce{HNO2}$ solution ($\pu{K_a= 4.5 \times 10^{-4}}$) calculate the concentration of the $\ce{H+}$ and $\ce{NO2-}$ ions, $\ce{HNO2}$ molecules and the $\ce{pH}$ of the solution.

If anyone could give any hints or answers that would be great.

## closed as off-topic by Pritt Balagopal, Jon Custer, Todd Minehardt, airhuff, jerepierreJun 14 '17 at 14:36

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• Pinpoint exactly where your stuck, otherwise this question will get closed as no effort. – Pritt Balagopal Jun 14 '17 at 10:54
• I'm mainly stuck on how to approach this question; I'm unsure how I can find the concentration of the H+ ion given this information. With this information I should be able to do this question, but I'm stumped on how I can find this value. Do you have any tips? :S – Pixel Rain Jun 14 '17 at 10:56
• I'm afraid you lack a few basic concepts before even attempting this question. Have you read any textbook regarding ionic equilibrium? Do you know what $K_a$ means? – Pritt Balagopal Jun 14 '17 at 10:58
• Isn't it the equilibrium constant for the reaction $H^+ + NO_2^- <-> HNO_2$? – Pixel Rain Jun 14 '17 at 11:03
• Precisely! Suppose if I give you a reaction: $$\ce{mA + nB <=>[K_a] pC + qD}$$ what would be the expression for $K_a$? – Pritt Balagopal Jun 14 '17 at 11:06

## 2 Answers

Answer:

$K_a = 4.5*10^-4$ Concentration (From ICE Table) of products/reactants:

HNO2 = 0.2 - x H+ = x NO2 = x

Therefore:

$$4.5*10^-4 = x^2/(0.2-x)$$ Rearrange: $$x^2 + x*(4.5*10^-4) - (0.2(4.5*10^-4)) = 0$$ Using quadratic formula: $x \approx 0.009$

$$pH = -log(10)$$ $$pH \approx 2.05$$.

That should be correct!

In order to tackle the problem, some assumptions have to be made first as seen in the working below which I have drawn up: In ideal circumstances, we assume that when one mole of nitrous acid dissociates it produces equimolar amounts of hydrogen ions and the anion of the acid. Thereby explaining assumption (I). Furthermore, as weak acids only dissociate slightly we can say that there will be negligible difference between the initial and equilibrium concentrations. It must be remembered that when dealing with these problems all concentrations must be from the solution at equilibrium. Because of assumption (I), we can represent the hydrogen ions and nitrous acid anions both as "x", which solves the problem you were having with the two unknowns. From then onwards all that has to be done is to solve for "x", which in the above solution is taken to two decimal places. But also keep in mind that the units have to be manipulated as well.

I believe you have forgot to add units for the K$_a$ value, as if you had then your concentration of "x" would come to mol$^{1/2}$dm$^{-3/2}$ which is not a valid value to be used in the pH equation as it requires an answer in moldm$^{-3}$. I hope this explanation furthers your progress.

• Welcome to SE! Nice to see your contribution. Try to avoid inserting picture texts, they're hard to search via googling. Also take a look at our guide to MathJax, to format math/chemistry expressions. – Pritt Balagopal Jun 15 '17 at 8:22