Question

a)a student tested a solution with a ph meter and recorded the ph to be 7.43. calculate the [H3O+] and [OH-] for this solution

b)a 2.291g sample containing sulfate ions was treated with barium to precipitate barium sulfate (BaSO4; MW=233.4g/mol) if the collected precipitate weighed 3.0687g was the unknown sample sodium sulfate (Na2SO4; MW-142.0g/mol) or potassium sulfate (K2SO4; MW=174.2g/mol) support your answer with calculations

c)consider the balanced reaction between Cu and nitric acid shown below. If 1.55g of Cu (NO3)2 (mw=187.6g/mol) was obtained from the reaction of 0.95g of Cu(mw=63.55g/mol) excess HNO3 what is the % yield of the reaction?
Cu(s) +4HNO3(aq)-> Cu(NO3)2(aq)+2NO2(g)+2H2O(l)

Answer 1

Answer on Question #56493 – Chemistry - General Chemistry

Question:

a) a student tested a solution with a pH meter and recorded the pH to be 7.43. calculate the $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$ and $\left[\mathrm{OH}^{-}\right]$ for this solution.

b) a 2.291 g sample containing sulfate ions was treated with barium to precipitate barium sulfate $\left(\mathrm{BaSO}_{4}; \mathrm{MW}=233.4 \mathrm{~g} / \mathrm{mol}\right)$. If the collected precipitate weighed 3.0687 g. Was the unknown sample sodium sulfate $\left(\mathrm{Na}_{2} \mathrm{SO}_{4}; \mathrm{MW}=142.0 \mathrm{~g} / \mathrm{mol}\right)$ or potassium sulfate $\left(\mathrm{K}_{2} \mathrm{SO}_{4}; \mathrm{MW}=174.2 \mathrm{~g} / \mathrm{mol}\right)$. Support your answer with calculations

c) consider the balanced reaction between Cu and nitric acid shown below. If $1.55\mathrm{g}$ of $\mathrm{Cu(NO_3)_2}$ (mw=187.6 g/mol) was obtained from the reaction of $0.95\mathrm{g}$ of $\mathrm{Cu(mw=63.55g/mol)}$ excess $\mathrm{HNO}_3$ what is the $\%$ yield of the reaction?

$$\mathrm{Cu}_{(s)} + 4 \mathrm{HNO}_{3(aq)} \rightarrow \mathrm{Cu(NO_3)_2(aq)} + 2 \mathrm{NO}_{2(g)} + 2 \mathrm{H_2O}_{(l)}$$

Answer:

a) $\mathrm{pH} = -\lg \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$

$\mathrm{pOH} = -\lg [\mathrm{OH}^{-}]$

$\mathrm{K_w = [H_3O^+][OH^-] = 10^{-14}}$

$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] + \left[\mathrm{OH}^{-}\right] = 14$

$\mathrm{pOH} = 14 - 7.43 = 6.57$

$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 10^{-pH}$

$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] = 10^{-7.43} = 3.71 \cdot 10^{-8} \mathrm{M}$

$\left[\mathrm{OH}^{-}\right] = 10^{-6.57} = 2.69 \cdot 10^{-7} \mathrm{M}$

b) $\mathrm{Ba}^{++} + \mathrm{K_2SO_4 = BaSO_4} + 2\mathrm{K}^+$

$\mathrm{Ba}^{++} + \mathrm{Na_2SO_4 = BaSO_4} + 2\mathrm{Na}^+$

According to the reaction, $v(BaSO_4) = v(K_2SO_4)$

$$v(BaSO_4) = v(Na_2SO_4)$$ $$v(BaSO_4) = \frac{3.0687}{233.4} = 0.013 \mathrm{mol}$$ $$v(K_2SO_4) = \frac{2.291}{174.2} = 0.013 \mathrm{mol}$$ $$v(Na_2SO_4) = \frac{2.291}{142.0} = 0.016 \mathrm{mol}$$

So that the unknown sample is Potassium Sulphate.

c) $v = \frac{m}{M}$

$$v(\mathrm{Cu}) = \frac{0.95}{63.55} = 0.02 \mathrm{mol}$$

According to the reaction, $v(\mathrm{Cu}) = v(\mathrm{Cu}(NO_3)_2)$

$$m(\mathrm{Cu}(NO_3)_2) = 0.02 \cdot 187.6 = 3.75 \mathrm{g}$$ $$\%(\mathrm{Cu}(NO_3)_2) = \frac{1.55}{3.75} \cdot 100 = 41.33\%$$

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