Strong acids show lower pH values while strong bases show higher pH values. Strong acids reacts with strong bases and give salts and water as products. When a strong base is added to a strong acid, pH value of acidic solution is increased. Also when a strong acid is added to a strong base, pH value of strong base solution is decreased.

**Content**

**Examples for strong acids****Examples for strong bases****pH equation****pOH equation****pH + pOH = pKw**- pH + pOH = pKw for a neutralized solution at 25
^{0}C<

- pH + pOH = pKw for a neutralized solution at 25
**Calculate pH of strong acids solutions**- Calculate pH of Hydrochloric acid solution
- Calculate pH of Sulfuric acid solution

**Calculate pH of strong bases solutions**- Calculate pH of Sodium hydroxide solution
- Calculate pH of Barium hydroxide solution

- Reaction of strong acids and strong bases

Refer following tutorial to how pH value of the acidic solution is changed when a strong base is added to the strong acid solution

Hydrochloric (HCl), sulfuric acid
(H_{2}SO_{4}), nitric acid
(HNO_{3} are some examples to strong acids. These acids completely hydrolysis to hydronium ions and respective
anions. Therefore, Hydronium ion concentration is higher than a Hydronium ion concentration of a weak acid.

Sodium hydroxide (NaOH), Potassium hydroxide (KOH), Barium hydroxide several are strong bases.

In this lesson, first we discuss about how to calculate pH of acidic solutions of strong acids such as HCl, H_{2}SO_{4}. Then,
we can find pH values of bases. Finally, we study reactions of strong acids and strong bases and calculating pH of these solutions
in different situations.

Hydronium ion (H_{3}O^{+}) concentration should be substituted in mol dm^{-3} to the pH equation.
Sometimes, we can represent H_{3}O^{+}_{(aq)} as H^{+}_{(aq)} and revised equation can be
presented as below.

**pH = -log _{10}[H^{+}_{(aq)}]**

Hydroxyl ion (OH^{-}) concentration should be substituted in mol dm^{-3} to the pOH equation.

Summation of pH and pOH value is a constant in a constant temperature. In 25^{0}C, pKa value is 14.

- In a neutralized aqueous solution, H
_{3}O^{+}concentration and OH^{-}concentration is equal to the 1 * 10^{-7}mol dm^{-3}. - Substitute those values in pH and pOH equation. You will get both pH and pOH values as 7. Therefore summation of pH and pOH values will be fourteen.

Example 1

You are provided 100 cm^{3} of HCl acid solution. It's concentration is 0.1 mol dm^{-3}.

- What is the concentration of H
^{+}? - Calculate the amount of moles of H
_{2}SO_{4}

**Hydrochloric acid completely dissociates to hydronium ions and
chloride ions in the water.**

HCl → H^{+}_{(aq)} + Cl^{-}_{(aq)}

According to the stoichiometry ratio, concentrations of,

[HCl] = [H_{3}O^{+}_{(aq)}], Therefore,

[HCl] = 0.1 mol dm^{-3}

Now, we know the concentration of H^{+}_{(aq)}. Let's substitute that value to the pH equation.

pH = -log_{10}[H_{3}O^{+}_{(aq)}]

pH = -log_{10}[0.1]

pH = 1.0

Example 2

You are provided 100 cm^{3} of H_{2}SO_{4} acid solution. It's concentration is 0.01 mol dm^{-3}.

- What is the concentration of H
^{+}? - Calculate the amount of moles of H
_{2}SO_{4}

H_{2}SO_{4(aq)} → 2H^{+}_{(aq)} + SO_{4}^{2-}_{(aq)}

According to the stoichiometry ratio, concentrations of,

2*[H_{2}SO_{4(aq)}] = [H_{3}O^{+}_{(aq)}], Therefore,

[H_{3}O^{+}_{(aq)}] = 2 * 0.01

[H_{3}O^{+}_{(aq)}] = 0.02 mol dm^{-3}

Now, we know the concentration of H^{+}_{(aq)}. Let's substitute that value to the pH equation.

pH = -log_{10}[H_{3}O^{+}_{(aq)}]

pH = -log_{10}[0.02]

pH = 1.7

Strong bases completely dissociates to ions in the water. Therefore, there is higher hydroxyl ion concentration in the aqueous solution. Aqueous strong base solutions show higher pH values.

- Calculating pOH of strong base
- Calculating pH by using pH + pOH = pKa

pH value of a strong base is found in two steps as mentioned above.

- Sodium hydroxide completely dissociates to Sodium cations and hydroxyl anions as following equation.
**NaOH**_{(aq)}→ Na^{+}_{(aq)}+ OH^{-}_{(aq)}- According to the stoichiometric balanced equation of dissociation, hydroxyl ion concentration is same as the to the initial concentration of base.
- Therefore, OH
^{-}concentration is 0.1 mol dm^{-3}. - Substitute OH
^{-}concentration to the pOH equation- pOH = -log
_{10}[OH^{-}_{(aq)}] - pOH = -log
_{10}(0.1) - pOH = 1

- pOH = -log

Because, pOH value is found in the last step and pKw value is 14 at 25^{0}C,

- pH + pOH = pKw
- pH + 1 = 14
- pH = 13

Therefore, 0.1 mol dm^{-3} NaOH solution is a strong basic solution.

here, we try to calculate pH value when a strong acid is mixed with a strong base or vice versa.

- Is final solution acidic or basic?
- pH of final solution?

An acid and a base(alkali) react and give salt and water as products. pH of acidic solution increases
when pouring a base into the acidic solution, and pH of basic solution decreases when pouring an acid into the basic solution.
This phenomenon happens due to **Neutralization** of acid or base.

HCl is a strong acid. NaOH is a strong base.Hence they react until one reactant finishes. Therefore, we have to find what reactant will remain.

- Calculate amount of moles(n) of each solution.
- Write the reaction and balance it.
- Find the ratio that reactants react.
- Find the reactant which remains in the solution. (HCl or NaOH)
- Calculate the concentration of remaining reactant.
- Calculate pH.

The relationship of dissolved moles, concentration and volume of solution can be given like this.

for HCl,

- amount of HCl = 0.1 mol dm
^{-3}* 50/1000 dm^{3}

- amount of HCl = 0.005 mol

for NaOH,

- amount of NaOH = 0.05 mol dm
^{-3}* 50/1000 dm^{3} - amount of NaOH = 0.0025 mol

NaOH and HCl react and give NaCl and H_{2}O as products. NaOH and HCl reacts 1:1 ratio. Therefore same amount of
moles of NaOH and HCl will react. Excess reactant(excess moles) will remain in the solution.

Construct a simple table which includes total initial moles of each reactant, number of moles
reacting(reactants) and producing(products). Then you can calculate remaining reactant. In this example 0.0025 moles of HCl
remains in the solution and solution becomes acidic. We don't consider about producing water because reaction occurs in **aqueous medium** and concentration
of water is very **high** and **don't change** due to reaction(change is negligible).

We know how many moles of HCl remaining in the solution now (0.0025 mol) and total volume of
the solution (100ml). So, we can calculate the concentration of HCl from concentration equation.

HCl is a strong acid. Therefore it dissociates completely into H^{+}
(or H_{3}O^{+}) and Cl^{-} in water. When we write the dissociation equation, we can see the
concentration of H_{3}O^{+} receives from HCl dissociation is equal to the concentration of HCl.

**Questions**

There are 0.05 moles of sulfuric acid 1 dm^{3} of aqueous solution. To calculate pH, you should know H^{+} concentration. Sulfuric acid is a strong acid and completely dissociates to H^{+} ions SO_{4}^{2-} ions.

So, if there are 0.05 moles of sulfuric, there are 0.01 moles of H^{+} ions. Volume of solution is 1 dm^{3}. Then concentration of H^{+} is 0.1 mol dm^{-3}.

Then pH is 1.

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