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Dennis_Churaev [7]
3 years ago
11

What is the concentration of a solution in which 15 grams of sugar is dissolved in 0.2 L of water?

Chemistry
1 answer:
Margaret [11]3 years ago
7 0

Answer:

0.2 M

Explanation:

Step 1: Given data

  • Mass of sugar (sucrose): 15 g
  • Volume of water: 0.2 L (we will assume it is the volume of the solution)

There are different ways to express the concentration of a solution. We will calculate molarity, which is one of the most used.

Step 2: Calculate the moles of sucrose

The molar mass of sucrose is 342.3 g/mol.

15 g × 1 mol/342.3 g = 0.044 mol

Step 3: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.044 mol/0.2 L = 0.2 M

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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
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In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

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