Question

A titration of 15.0 cm3 of household ammonia, NH3, required 3870 cm3 of 8.0 M HCl. Calculate the molarity of the ammonia

Answer 1

Answer on Question #50479, Chemistry, Other

A titration of $15.0~\mathrm{cm}^3$ of household ammonia, $\mathrm{NH}_3$, required $3870~\mathrm{cm}^3$ of $8.0~\mathrm{M}$ HCl. Calculate the molarity of the ammonia.

Solution:

$$\mathrm{NH}_3 + \mathrm{HCl} \rightarrow \mathrm{NH}_4\mathrm{Cl}$$ $$n = C \times V$$ $$n(NH_3) = n(HCl)$$ $$C_1 \times V_1 = C_2 \times V_2$$ $$C_2 = \frac{C_1 \times V_1}{V_2} = \frac{3870~\mathrm{cm}^3 \times 8.0~\mathrm{M}}{15~\mathrm{cm}^3} = 2064~\mathrm{M}$$

Answer:

**2064 M it is impossible to have such concentration!** In Question is a mistake.

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