In chemistry, aqueous solutions are prepared for various purposes. As examples, acidic or basic aqueous solutions are prepared by dissolve a specific quantity of chemical compound. According to the concentration and volume of the required solution, weight or volume of solid chemical or concentrated solution should be measured or calculated. As well, pH value of the solution is given to calculate concentration and other parameters.

In this tutorial, we are going to practice some problems which contains relationships of mass, concentration, volume and pH in aqueous solutions. You can understand how to apply those equations in such problems to solve these type of questions easily.

**Content**

- Calculate required mass of KOH to prepare 25cm
^{3}of 0.1 mol dm^{-3}KOH solution. - Calculate the mass of NaOH required to prepare 500ml of NaOH solution of pH = 13 at is
25
^{0}C temperature.

**Question 1**

Before solve this kind of equation, following steps should be followed by you.

- Draw a sketch to mark known and unknown data
- Build a relation of equations to calculate unknown parameters.
- Step 1: Volume (V) and concentration (C) of the solution are given as data. Therefore amount of solute (n) can be calculated using the C = n/V equation.
- Step 2: Use n = m/M equation to calculate required mass of KOH solid. M = molar mass of compound (in this case KOH)

- C = n/V
- n = C × V
- Required amount of KOH = 0.1 mol dm
^{-3}× 25 × 10^{-3}dm^{3} - Required amount of KOH = 0.0025 mol

In step 1, amount of KOH was calculated. Now, we can use the relationship of mass (m), molar mass (M) and amount (n) to calculate required mass of KOH. As an additional data, molar mass of molar mass of KOH should be calculated prior to calculate mass of KOH.

- Molar mass of KOH = 39 + 16 + 1
- Molar mass of KOH = 56 g mol
^{-1}

Then apply n = m/M equation.

- n = m/M
- m = n × M
- Required mass of KOH = 0.0025 mol × 56 g mol
^{-1} - Required mass of KOH = 0.14 g

**Question 2**

Calculate pOH from pH by using
the **pH + pOH = 14 **(at 25^{0}C)

Then you can calculate the concentration of NaOH solution by **pOH = -log _{10}[OH^{-}_{(aq)}]** .

Then find the required moles of NaOH by the equation of

Now weight is measured by multiplying number of moles and molar mass.

**Assumptions**

The OH^{-} received from water dissociation is negligible when it compares with OH^{-} received
from NaOH.

pH + pOH = 14

13 + pOH = 14

pOH = 1

Now we know the pOH of NaOH solution. Use the relation of pOH and OH^{-} concentration to calculate the concentration of OH^{-}.

pOH = -log_{10}[OH^{-}_{(aq)}]

1 = -log_{10}[OH^{-}_{(aq)}]

[OH^{-}_{(aq)}] = 0.1 mol dm^{-3}

Concentration, moles amount, and volume relation is used.

C = n/v

0.1 = n/0.5 [500 ml = 0.5 dm^{3}]

n = 0.05mol

NaOH amount = 0.05 mol

Multiply required moles times molecular weight

m = mass

n = moles

M = molecular weight

n = m/M

m = molecular weight * moles

m = M*n

m = 0.05 * 40

m = 2g

Ok. First we should know, what are the known parameters of the given solution. In this example, pH and volume are given as data.

When pH value is known, concentration of H_{3}O_{+} or OH_{-} can be foundfrom pH or pOH equation.

According to the H_{3}O_{+} or OH_{-} concentration and stoichiometric ratio of dissociation of the compound, the concentration of dissolved compound is found.

With found concentration and volume of solution, dissolved amount is found.

Amount = concentration * volume

Next, we use relationship of amount, mass and molecular mass.

Amount = mass / molecular mass

Because we know the amount and molecular mass, dissolved mass can be found.

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