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?