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CHEMISTRY, M5 2019 HSC 17 MC

A student makes a solution with a final volume of 200 mL by mixing 100 mL of 0.0500 mol L ¯1 barium nitrate solution with 100 mL of 0.100 mol L ¯1 sodium hydroxide solution.

Which row of the table correctly identifies if a precipitate will form under these conditions and the reason?
 

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`D`

Show Worked Solution

\(\ce{Ba(OH)2(s) \rightleftharpoons Ba^2+(aq) + 2OH-(aq)}\)

`K_(sp)=2.55xx10^(-4)`

\[\ce{[Ba^2+] = \frac{(0.05)(0.1)}{(0.2)} = 0.025 M}\]

\[\ce{[OH-] = \frac{(0.1)(0.1)}{(0.2)} = 0.05 M}\]

`Q` \(\ce{= [Ba^2+][OH-]^2}\)  
  `=(0.025)(0.05)^2`  
  `=6.25xx10^(-6)`  

 
Since `Q<K_(sp)` , no precipitate forms.

`=>D`


♦♦♦ Mean mark 36%.

CHEMISTRY, M6 2020 HSC 33

Excess solid calcium hydroxide is added to a beaker containing 0.100 L of 2.00 mol L¯1 hydrochloric acid and the mixture is allowed to come to equilibrium.

  1. Show that the amount (in mol) of calcium hydroxide that reacts with the hydrochloric acid is 0.100 mol.   (2 marks)

  2. It is valid in this instance to make the simplifying assumption that the amount of calcium ions present at equilibrium is equal to the amount generated in the reaction in part (a).
  3. Calculate the pH of the resulting solution.  (4 marks)

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a.   \(\ce{Ca(OH)2 (s) + 2 HCl (aq) -> CaCl2 (aq) + 2 H2O (l)}\)

`text{n(HCl)} = text{c} xx text{V} = 2.00 xx 0.100  = 0.200\ text{mol}`

\[\ce{n(Ca(OH)2) = \frac{n(HCL)}{2} = \frac{0.200}{2}= 0.100 mol}\]  

b.   `text{pH} = 11.35`

Show Worked Solution

a.   \(\ce{Ca(OH)2 (s) + 2 HCl (aq) -> CaCl2 (aq) + 2 H2O (l)}\)

`text{n(HCl)} = text{c} xx text{V} = 2.00 xx 0.100  = 0.200\ text{mol}`

\[\ce{n(Ca(OH)2) = \frac{n(HCL)}{2} = \frac{0.200}{2}= 0.100 mol}\]  

b.   \(\ce{Ca(OH)2(s) \rightleftharpoons Ca^2+ (aq) + 2 OH– (aq)}\)

`[text{Ca}^(2+)] = text{n} / text{V} = 0.100 / 0.100 = 1.00\ text{mol L}^-1`
 

\(\ce{K_{sp}}\) \( \ce{= [Ca^2+][OH– ]^2}\)  
`5.02 xx 10^(-6)` `= 1.00 xx [text{OH}^– ]^2`  
`[text{OH}^– ]` `=sqrt{5.02 xx 10^(-6)}`  
  `=2.24 xx 10^(−3)\ text{mol L}^(-1)`  
`text{pOH }` `= −log_10(2.24 xx 10^(-3))`  
  `= 2.650`  

 
`:.\ text{pH} = 14-2.650 = 11.35`


♦♦♦ Mean mark (b) 20%.

CHEMISTRY, M5 2021 HSC 19 MC

A quantity of silver nitrate is added to 250.0 mL of 0.100 mol L ¯1 potassium sulfate at 298 K in order to produce a precipitate. Silver nitrate has a molar mass of 169.9 g mol ¯1.

What mass of silver nitrate will cause precipitation to start?

  1. 0.00510 g
  2. 0.186 g
  3. 0.465 g
  4. 0.854 g
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`C`

Show Worked Solution

The reaction when silver nitrate is added to potassium sulfate is:

`text{2} text{Ag} text{NO}_(3) (aq) + text{K}_(2) text{SO}_(4) (aq) →  text{Ag}_(2) text{SO}_(4) (s) + text{2} text{K} text{NO}_(3) (aq)`
 

Since each `text{K}_(2) text{SO}_(4)` molecule has 1 sulfate ion

`[text{SO}_(4) ^(2-)] = text{K}_(2) text{SO}_(4) = 0.100\ text{mol L}^(-1)`
 

`text{Ag}_(2) text{SO}_(4) (s) ⇋ 2text{Ag}^(+) (aq) + text{SO}_(4) ^(\ 2-) (aq)`

`text{K}_(sp) = [text{Ag}^(+)]^2 [text{SO}_(4) ^(2-)]`
 

From the data sheet:

`text{K}_(sp) = text{1.20 x 10}^-5`

`text{1.20 x 10}^-5 = [text{Ag}^(+)]^2  xx  [text{SO}_(4) ^(\ 2-)]`

`text{1.20 x 10}^-5 = [text{Ag}^(+)]^2  xx  [text{0.100}]`

`[text{Ag}^(+)] = 0.01095…\ text{mol L}^(-1)`

`[text{Ag}^(+)] = [text{AgNO}_(3)]`
 

`text{n(AgNO}_(3)) = text{c}  xx text{V} = 0.01095… xx 0.250 = 0.00273…\ text{mol}`

`text{m(AgNO}_(3)) = text{n} xx text{MM} = 0.00273… xx 169.9 = 0.465\ text{g}`

`=> C`


♦ Mean mark 45%.