Question

Why are hydrogen ions NEVER found in an aqueous solution?


HCN(aq) + SO4-2(aq) HSO4-(aq) + CN -(aq)
What is the Bronsted - Lowry acid in this equation?
What is the Bronsted - Lowry base in this equation?
What is the conjugate acid in this equation?
What is the conjugate base in this equation?


2NH3 + Ag+ Ag(NH3)2+
What is the Lewis acid in this equation?
What is the Lewis base in this equation?


Part II

Given H2SO4 is sulfuric acid, HNO3 is nitric acid, and H3PO4 is phosphoric acid, name the following:
HCl
H2SO3
HNO2
H3PO2
HNO4
H2SO5
HI
Part III

Work these titration calculations:



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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
Calculate the volume of 0.55M HNO3 necessary to neutralize 5500 cm3 of 0.45M KOH.

Answer 1

Answer on Question #50438, Chemistry, Other

Task:

Part I

a) Why are hydrogen ions NEVER found in an aqueous solution?

$\mathrm{HCN(aq) + SO_4^{2}(aq)\rightarrow HSO_4^{-}(aq) + CN^{-}(aq)}$

What is the Bronsted - Lowry acid in this equation?

What is the Bronsted - Lowry base in this equation?

What is the conjugate acid in this equation?

What is the conjugate base in this equation?

b) $2 \mathrm{NH}_{3} + \mathrm{Ag}^{+} \rightarrow \mathrm{Ag}(\mathrm{NH}_{3})_{2}^{+}$

What is the Lewis acid in this equation?

What is the Lewis base in this equation?

Part II

Given $\mathrm{H}_2\mathrm{SO}_4$ is sulfuric acid, $\mathrm{HNO}_3$ is nitric acid, and $\mathrm{H}_3\mathrm{PO}_4$ is phosphoric acid, name the following: HCl, $\mathrm{H}_2\mathrm{SO}_3$, $\mathrm{HNO}_2$, $\mathrm{H}_3\mathrm{PO}_2$, $\mathrm{HNO}_4$, $\mathrm{H}_2\mathrm{SO}_5$, HI.

Part III

Work these titration calculations:

a) A titration of $15.0~\mathrm{cm}^3$ of household ammonia, $\mathrm{NH}_3$, required $38.70~\mathrm{cm}^3$ of 8.0 M HCl. Calculate the molarity of the ammonia.

b) What volume of 5.0 M HNO₃ is required to neutralize $25.00~\mathrm{cm}^3$ of a 2.0 M NaOH solution.

c) Calculate the volume of 0.55 M HNO₃ necessary to neutralize $55.00~\mathrm{cm}^3$ of 0.45 M KOH.

Answer:

Part I

a) Hydrogen ions never found in an aqueous solution because the attach to the water molecules to form $\mathrm{H}_3\mathrm{O}^+$.

Bronsted - Lowry acid in the equation is HCN(aq), because it donates the proton.

Bronsted - Lowry base in this equation is $\mathrm{SO}_4^{2-}(aq)$, because it receives the proton.

Conjugate acid in this equation is $\mathrm{SO}_4^{2-}(aq)$.

Conjugate base in this equation is HCN(aq).

b) Lewis acid in this equation is $\mathrm{Ag}^{+}$.

Lewis base in this equation is $\mathrm{NH}_3$.

Part II

HCl – hydrochloric acid; $\mathrm{HNO}_4$ – peroxynitric acid;

$\mathrm{H}_2\mathrm{SO}_3$ – sulforous acid; $\mathrm{H}_2\mathrm{SO}_5$ – peroxymonosulphuric acid;

$\mathrm{HNO}_2$ – nitrous acid; HI – hydroiodic acid.

$\mathrm{H}_3\mathrm{PO}_2$ – hypophosphorous acid;

Part III

a)

$$NH_4OH + HCl = NH_4Cl + H_2O$$ $$C_M = \frac{\nu}{V} \quad \nu = C_M \cdot V$$ $$\nu(NH_4OH) = \nu(HCl)$$ $$C_M(NH_4OH) \cdot V(NH_4OH) = C_M(HCl) \cdot V(HCl)$$ $$C_M(NH_4OH) = \frac{C_M(HCl) \cdot V(HCl)}{V(NH_4OH)}$$ $$C_M(NH_4OH) = \frac{8 \cdot 0.0387}{0.015} = 20.64\,M$$ $$HNO_3 + NaOH = NaNO_3 + H_2O$$ $$C_M = \frac{\nu}{V} \quad \nu = C_M \cdot V$$ $$\nu(HNO_3) = \nu(NaOH)$$

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

$$V(HNO_3) = \frac{C_M(NaOH) \cdot V(NaOH)}{C_M(HNO_3)}$$ $$V(HNO_3) = \frac{2 \cdot 0.025}{5} = 0.01\,l = 10\,cm^3$$ $$HNO_3 + KOH = KNO_3 + H_2O$$ $$\nu(HNO_3) = \nu(KOH)$$

c) $C_M(HNO_3) \cdot V(HNO_3) = C_M(KOH) \cdot V(KOH)$

$$V(HNO_3) = \frac{C_M(KOH) \cdot V(KOH)}{C_M(HNO_3)}$$ $$V(HNO_3) = \frac{0.45 \cdot 0.055}{0.55} = 0.045\,l = 45\,cm^3$$

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