Question

How do we solve this question? I found B answer key says A

Answer 1

First, we write the reaction and balance it:

HNO2 (aq) + NH3 (aq) = NH4+ + NO2- (Balanced)

Data:

50 mL of 0.2 M HNO2

50 mL of 0.2 M NH3

In total, we have 100 mL, therefore, this solution between HNO2 and NH3 will be diluted in half. I mean: The concentration of HNO2 and NH3 will be 0.10 M

HNO2 (aq) + NH3 (aq) = NH4+ + NO2-

Initial 0.10 M 0.10 M 0 0

reacts -x -x +x +x

Equilibrium 0.10-x 0.10-x +x +x

Now, we write Kc:

`\begin{gathered} Kc\text{ = }(\lbrack NH4+\rbrack\lbrack NO2-\rbrack)/(\lbrack HNO2\rbrack\lbrack NH3\rbrack)=(x^2)/((0.10-x)^2) \\ 1x10^6=(x^2)/((0.10-x)^2) \\ √(1x10^6)=\text{ }\lvert{(x)/((0.10-x))}\rvert \\ We\text{ get 2 values here:} \\ 1)+1000=(x)/((0.10-x)) \\ and \\ 2)-1000\text{ = }(x)/((0.10-x)) \end{gathered}`

Values of x:

For 1) x = 0.0999

For 2)x = 0.1001

We choose number 1) x = 0.0999

Number 2 gives us a value higher than the initial values of concentration

Therefore, concentration in equilibrium of NH3 = 0.10-x =0.10 - 0.0999 = 0.00010M

Answer: A. 0.00010M