65 g of NaOH is dissolved to make 500 mL of solution. Find moles of NaPH and Molarity of solution.

Chemistry

1 answer:

Lisa [10]2 years ago

8 0

Answer:

3.25 M

Explanation:

Number of moles = mass/ molar mass

mass of NaOH = 65 g

Molar mass of NaOH = 40 g/mol

Number of moles of NaOH = 65g/40 g/mol = 1.625 moles

Let us recall that;

n = CV

n = number of moles of solute

C = concentration of solution

V = volume of solution

C = n/V

n = 1.625 moles

V = 500/1000 L = 0.5 L

C = 1.625 moles/0.5

C = 3.25 M

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The <em>balanced equation</em> would be:

FeCl₃ (aq) + 3 NH₄OH (aq) → Fe(OH)₃ (s) + 3 NH₄Cl (aq)
The precipitate is Fe(OH)₃ or iron iii hydroxide.

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EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.