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 #50367, Chemistry, Other

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

Solution:

$$n = c \times V$$ $$\mathrm{NH_3 \times H_2O + HCl \rightarrow NH_4Cl + H_2O}$$ $$\begin{array}{l} n_1 = n_2 \\ c_1 \times V_1 = c_2 \times V_2 \\ c_2 = \frac{c_1 \times V_1}{V_2} \\ \end{array}$$ $$c_{NH_3} = \frac{8.0~M \times 3870~\mathrm{cm^3}}{15~\mathrm{cm^3}} = 2064~\mathrm{M}$$

Answer:

molarity of the ammonia is 2064 M

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