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alekssr [168]
4 years ago
6

Wastewater discharged into a stream by a sugar refinery contains 1.70 g of sucrose (C12H22O11) per liter. A government-industry

project is designed to test the feasibility of removing the sugar by reverse osmosis. What pressure must be applied to the apparatus at 20.°C to produce pure water?
Chemistry
1 answer:
erik [133]4 years ago
5 0

Answer:

A pressure value more than 0.012 atm must be applied to the apparatus at 20.0°C to produce pure water.

Explanation:

When pressure more than the value of osmotic pressure of the solution is applied to the solution then water will move out from the solution which is termed as reverse osmosis.

Given that 1.70 grams of sucrose was present in 1 L of solution.

Moles of sucrose present in 1 liter of solution  = \frac{1.70 g}{342 g/mol}=0.004971 mol/L

C = 0.004971 mol/L

Temperature of the solution = T = 20.0°C = 20.0+273 K = 293 K

Osmotic pressure of soluion = \Pi

\Pi=CRT

\Pi=0.004971 mol/L\times 0.0821 atm L/mol K\times 293 K=0.012 atm

A pressure value more than 0.012 atm must be applied to the apparatus at 20.0°C to produce pure water.

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Consider the titration of 100.0 mL of 0.100 M acetic acid with 0.100 M NaOH. CH3CO2H(aq) + OH-(aq) → CH3CO2-(aq) + H2O(ℓ) Ka for
allsm [11]

Answer:

a) pH = 5.70

b) pH = 8.22

c)pH = 11.68

Explanation:

a)NaOH (aq)+CH3COOH(aq) ------> CH3COONa(aq) + H2O(aq)

1mol of NaOH react with 1mol of CH3COOH

No of mole in 90ml of 0.1M NaOH

                       = (0.1mol/1000ml)×90ml

                       = 0.009

No of mol in 100ml of CH3COOH

                     = (0.1/1000)×100

                     = 0.01

No of mol of CH3COOH after addition

                    = 0.01-0.009

                    = 0.001

Total volume = 100ml + 90ml

                     = 190ml

Final molarity of CH3COOH =( 0.0025/190) ×1000

                    = 0.00526M

Concentration of CH3COONa formed =( 0.0075/190) ×1000

                      = 0.0474M

Ka of CH3COOH = 1.8 × 10^-5

pka = -log(ka)

pKa = 4.75

Applying Henderson equation

pH = pKa + log ( [A-]/[HA])

pH = 4.75 + log ( 0.0474/0.00526)

= 5.70

b)

At equivalencepoint point ,  

No of moles of CH3COOH = 0

No of moles of CH3COO- = 0.01 mol

Total volume = 200ml

molarity of CH3COO- = 0.01/2

                     = 0.0050M

CH3COO- (aq) + H2O(l) <---------> CH3COOH(aq) + OH-(aq)

Kb = [ CH3COOH] [ OH- ] / [ CH3COO- ]

                  = 1.8 × 10^-5

[ CH3COOH ] = X

[ OH-] = X

[ CH3COO-] = 0.0050 - X

5.6 × 10^-10 = X^2/ (0.0050 - X)

we can assume , 0.0050 - X = 0.0050

5.6 × 10^-10 = X^2/0.0050

X = 1.67 × 10^6

[OH-] = 1.67 × 10^-6

pOH = 5.78

pH = 14 - pOH

pH = 14 -5.78

pH = 8.22

c) No of mol of OH from excess 10ml of NaOH = (0.1mol /1000ml)×10ml = 0.001mol

No of mol of OH- from hydrolysis of CH3COO- = (1.67×10^-6/1000)×200= 3.34×10^-7mol

Second one is negligible

So, no of mol OH- = 0.001mol

Total volume = 100ml + 110ml

                = 210ml

[OH-] =( 0.001/210)×1000

      = 0.0048M

pOH = -log[OH-]

= - log (0.0048)  

= 2.32

pH = 14 - 2.32

pH = 11.68

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