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assuming no determinate errors, within what range are you at 95% confid…

Question

assuming no determinate errors, within what range are you at 95% confident
level that the true mean value of the calcium content falls.
(4 marks)
b) describe three major steps in a qualitative analysis of heavy metals in soil
samples.
(6 marks)
c) 0.470g of sample containing sodium carbonate was treated with 50ml of 0.10m
hcl. the excess hcl was back titrated using 0.1m standardized naoh were
20ml of naoh was used. calculate the percentage na₂co₃ in the sample. (rfm
of na₂co₃ = 105.99g/mol)
(5 marks)
d) calculate the ph of 4.0 x10⁻⁴m and 0.001m solutions of ethanoic acid given that
its kₐ= 1.8 x10⁻⁵
(5 marks)
question four
a) describe how the following solutions are prepared in the laboratory, show all your
workings:
i) 500ml of 0.6m hydrochloric acid (density of hcl = 1.18g/cm³,
h=1.007, cl= 35.456, %purity = 37%)
(3 marks)
ii) 400ml of 0.5m sodium carbonate na₂co₃
(2 marks)
iii) 500ml of 2.0m nitric acid (n=15.00, o = 15.999, %purity = 63%, density
of hno₃= 1.51g/cm³)
(2 marks)

Explanation:

Response
Part c)

Step 1: Determine moles of HCl initially

Molarity of HCl, \( M_{HCl} = 0.10 \, M \), volume of HCl, \( V_{HCl} = 50 \, mL = 0.05 \, L \).
Moles of HCl, \( n_{HCl} = M_{HCl} \times V_{HCl} = 0.10 \, mol/L \times 0.05 \, L = 0.005 \, mol \).

Step 2: Determine moles of NaOH used

Molarity of NaOH, \( M_{NaOH} = 0.1 \, M \), volume of NaOH, \( V_{NaOH} = 20 \, mL = 0.02 \, L \).
Moles of NaOH, \( n_{NaOH} = M_{NaOH} \times V_{NaOH} = 0.1 \, mol/L \times 0.02 \, L = 0.002 \, mol \).

Step 3: Moles of HCl reacted with \( Na_2CO_3 \)

HCl and NaOH react in 1:1 ratio (\( HCl + NaOH = NaCl + H_2O \)). So moles of HCl reacted with NaOH is 0.002 mol.
Moles of HCl reacted with \( Na_2CO_3 \), \( n_{HCl(reacted\ with\ Na_2CO_3)} = 0.005 - 0.002 = 0.003 \, mol \).

Step 4: Moles of \( Na_2CO_3 \)

Reaction: \( Na_2CO_3 + 2HCl = 2NaCl + H_2O + CO_2 \). So 1 mol \( Na_2CO_3 \) reacts with 2 mol HCl.
Moles of \( Na_2CO_3 \), \( n_{Na_2CO_3} = \frac{n_{HCl(reacted\ with\ Na_2CO_3)}}{2} = \frac{0.003}{2} = 0.0015 \, mol \).

Step 5: Mass of \( Na_2CO_3 \)

RFM of \( Na_2CO_3 = 105.99 \, g/mol \).
Mass of \( Na_2CO_3 \), \( m = n \times RFM = 0.0015 \, mol \times 105.99 \, g/mol = 0.158985 \, g \).

Step 6: Percentage of \( Na_2CO_3 \)

Percentage \( = \frac{mass\ of\ Na_2CO_3}{mass\ of\ sample} \times 100 = \frac{0.158985}{0.470} \times 100 \approx 33.83\% \).

Step 1: Calculate mass of HCl needed

Molarity \( M = 0.6 \, M \), volume \( V = 500 \, mL = 0.5 \, L \).
Moles of HCl, \( n = M \times V = 0.6 \, mol/L \times 0.5 \, L = 0.3 \, mol \).
Molar mass of HCl, \( M_{HCl} = 1 + 35.456 = 36.456 \, g/mol \).
Mass of pure HCl, \( m_{pure} = n \times M_{HCl} = 0.3 \times 36.456 = 10.9368 \, g \).

Step 2: Calculate mass of commercial HCl needed

Purity = 37%, so \( \text{Mass of commercial HCl} = \frac{m_{pure}}{\text{purity}} = \frac{10.9368}{0.37} \approx 29.56 \, g \).

Step 3: Calculate volume of commercial HCl

Density \(
ho = 1.18 \, g/cm^3 \), volume \( V = \frac{\text{mass}}{
ho} = \frac{29.56}{1.18} \approx 25.05 \, mL \).

Step 4: Preparation steps

  1. Measure ~25.05 mL of commercial HCl using a graduated cylinder.
  2. Transfer to a 500 mL volumetric flask.
  3. Add distilled water to the mark, mix well.

Step 1: Calculate moles of \( Na_2CO_3 \)

Molarity \( M = 0.5 \, M \), volume \( V = 400 \, mL = 0.4 \, L \).
Moles of \( Na_2CO_3 \), \( n = M \times V = 0.5 \times 0.4 = 0.2 \, mol \).

Step 2: Calculate mass of \( Na_2CO_3 \)

Molar mass of \( Na_2CO_3 = 105.99 \, g/mol \).
Mass, \( m = n \times M = 0.2 \times 105.99 = 21.198 \, g \).

Step 3: Preparation steps

  1. Weigh 21.20 g of \( Na_2CO_3 \) (anhydrous) using a balance.
  2. Transfer to a 400 mL volumetric flask (or dissolve in beaker, then transfer).
  3. Add distilled water to the mark, mix well.

Answer:

Approximately \( 33.83\% \)

Part d)

For ethanoic acid (\( CH_3COOH \)), \( K_a = 1.8 \times 10^{-5} \), formula for pH of weak acid: \( pH = \frac{1}{2}(pK_a - \log c) \), where \( c \) is concentration.

For \( c = 4.0 \times 10^{-3} \, M \):

Step 1: Calculate \( pK_a \)

\( pK_a = -\log K_a = -\log(1.8 \times 10^{-5}) \approx 4.74 \).

Step 2: Calculate pH

\( pH = \frac{1}{2}(4.74 - \log(4.0 \times 10^{-3})) \).
\( \log(4.0 \times 10^{-3}) = \log 4 + \log 10^{-3} \approx 0.602 - 3 = -2.398 \).
\( pH = \frac{1}{2}(4.74 - (-2.398)) = \frac{1}{2}(7.138) \approx 3.57 \).

For \( c = 0.001 \, M = 1.0 \times 10^{-3} \, M \):

Step 1: Calculate \( \log c \)

\( \log(1.0 \times 10^{-3}) = -3 \).

Step 2: Calculate pH

\( pH = \frac{1}{2}(4.74 - (-3)) = \frac{1}{2}(7.74) \approx 3.87 \).