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.
What volume of 5.0 M HNO3 is required to neutralize 2500 cm3 of a 2.0 M NaOH solution

Answer 1

Answer on Question #50450, Chemistry, Other

Task:

A titration of 15.0 cm³ of household ammonia, NH₃, required 38.70 cm³ of 8.0 M HCl. Calculate the molarity of the ammonia.

What volume of 5.0 M HNO₃ is required to neutralize 25.00 cm³ of a 2.0 M NaOH solution.

Answer:

$$NH_4OH + HCl = NH_4Cl + H_2O$$ $$C_M = \frac{v}{V} \quad v = C_M \cdot V$$ $$v(NH_4OH) = v(HCl)$$

1) $C_M(NH_4OH) \cdot V(NH_4OH) = C_M(HCl) \cdot V(HCl)$

$$C_M(NH_4OH) = \frac{C_M(HCl) \cdot V(HCl)}{V(NH_4OH)}$$ $$C_M(NH_4OH) = \frac{8 \cdot 0.0387}{0.015} = 20.64\,M$$ $$HNO_3 + NaOH = NaNO_3 + H_2O$$ $$C_M = \frac{v}{V} \quad v = C_M \cdot V$$ $$v(HNO_3) = v(NaOH)$$

2) $C_M(HNO_3) \cdot V(HNO_3) = C_M(NaOH) \cdot V(NaOH)$

$$V(HNO_3) = \frac{C_M(NaOH) \cdot V(NaOH)}{C_M(HNO_3)}$$ $$V(HNO_3) = \frac{2 \cdot 0.025}{5} = 0.01\,l = 10\,cm^3$$

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