How to find the concentration of an of ammonia solution by titration with nitric acid? I am stuck with this question I am doing:

To determine the ammonia ($\ce{NH3}$) concentration in a cleaning product, a chemist diluted 20 mL of the product to 500 mL. Of the 500 mL solution, 25 mL was titrated with 0.151 M $\ce{HNO3}$. The endpoint occurred at 26.2 mL. Calculate the concentration of the cleaning fluid.

[Ans: 3.956 M]

What I have done so far:


I worked out the moles of $\ce{HNO3}$ that the $\ce{NH3}$ must have reacted with, using the concentration and volume they gave me. I got 0.0039562 moles.
I then wrote the balanced equation and worked out that the stoichiometric ratio of $\ce{HNO3:NH3}$ must be 1:1. Therefore, I worked out that the amount of moles of $\ce{NH3}$ reacting with the $\ce{HNO3}$, must also be 0.0039562 moles.
From here, I initially worked out the concentration of $\ce{NH3}$ reacting with $\ce{HCl}$ and got 0.158 M, but then I realised this wasn't what the question was asking. I tried using 0.02 mL as the volume instead (as this was the very initial volume of $\ce{NH3}$) but still got the wrong answer.


Can someone please tell me what I have done wrong/ guide me in the right direction.

Edit: I have just tried it again, and this time I used the concentration I previously found (0.158 M) and a volume of 0.5 L to work out the number of moles in the 500 mL dilution of the product. I used this value which was 0.079 moles and a volume of 0.02 mL which was the original volume of cleaning liquid, to work out the concentration. I seem to be getting the right answer, but is my working correct?
Yes, although your presentation is a bit confusing.
To be clear,


You determined the quantity of $\ce{HNO3}$ consumed during the titration 
$$0.151\ \mathrm{M} \times 26.2\ \mathrm{mL} = 3.956 \  \mathrm{mmol}$$
Then given this titration equation:
$$\ce{HNO3 + NH3 -&gt; NO3- + NH4+}$$
You deduced that the $25~\mathrm{ml}$ sample contained $3.956 \  \mathrm{mmol}$ of $\ce{NH3}$, hence a concentration of 
$$\frac{3.956\ \mathrm{mmol}}{25\ \mathrm{mL}}=0.1582\ \mathrm{M}$$
From which you calculated the amount of $\ce{NH3}$ in the $500\ \mathrm{mL}$ sample:
$$0.1582\ \mathrm{M} \times 500\ \mathrm{mL} = 79.12 \  \mathrm{mmol}$$
And finally since the original $20\ \mathrm{mL}$ sample also contains $79.12\ \mathrm{mmol}$ of $\ce{NH3}$, that gives you the original concentration:
$$\frac{79.12\ \mathrm{mmol}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$


What I think you should have done in place of 4 and 5 is:

4'. Since the $25\ \mathrm{mL}$ sample is taken from a solution diluted from $20\ \mathrm{mL}$ to $500\ \mathrm{mL}$, the original concentration is:
$$0.1582 \ \mathrm{M} \times \frac{500\ \mathrm{mL}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$

Which spares you one calculation's work and potential errors.