The pH of a solution of 0.1 M NaOH is

You can calculate pH if you know the concentration of the NaOH.

You will need to know the molarity of the NaOH. Let's assume the solution is 0.1M. NaOH is a strong base, so this will produce 0.1mol/L of OH ions in solution. This will produce a pH of 13.

You will need to take the negative log of 0.1 to find the pOH. This will work out to be 1.

Since pH + pOH = 14

We can calculate the pH to be 13.

This assumption we made about the base can only be used for strong bases which dissociate completely in water.

Hope this helps!

Solved Problems

1. Calculate the pH of a 0.01 M HCl solution.

HCl is a strong electrolytes and thus its complete dissociation will produce a solution O.O1 M (10-2 M) in H+. Since the concentration of H+  arising from water dissociation (10-7 M) is notably lesser than 10-2 M, it can be neglected and thus

pH = -Log (10-2) = 2

2. Calculate the pH of a 0.1 M NaOH solution.

NaOH is a strong electrolytes and thus its complete dissociation will produce a solution O.1 M (10-2 M) in OH-. By neglecting [OH] arising from water dissociation (10-7 M)

[H+][OH-] = [H+][OH-]) = 10-14

from which [H+] = 10-13 and  pH = - log (10-13) = 13

Alternatively we can calculate pOH = - log [OH-]= - log [10-1] = 1 and remembering that pH + pOH = 14 we get pH = 14 - 1 = 13.

3. How many grams of NaOH should be dissolved in two liters of water to have pH = 10.7 ?

pH = - log [H+] = 10.7

log [H+] = 10.7

remembering the definition of logarithms

[H+] = 10-10.7

By using the expression for ion product of water:

[OH-][H+] = 10-14

[OH-] = 10-14/10-10.7 = 5.01-14 M

Since NaOH is a strong base we need to add 5.01-14 moles of NaOH for liter of water, 10.02-14 for two liters of water. By multiplying the moles with the molecular weight of NaOH we have

 (10.02-14) (40) = 0.04 grams

4. An HCl solution having pH = 3 should be diluted in such a way to obtain pH = 4. How many water you need to add?

pH = 3 ==> [H+] = 10-3 M

pH = 4 ==> [H+] = 10-4 M

remembering that M1V1 = M2V2

(10-3)(3) = (10-4)(x)

x = (10-3)(3)/(10-4) = 30 liters

Thus you need to add 27 liters of water.

Alternatively considering that a 10-4 M solution is ten fold more diluted than a 10-3 M, the final volume should be (3)(10) = 30

5. 300 ml of a 0.02 M NaOH solution are mixed with 200 ml of 10-2 M Ba(OH)2 solution. Calculate the pH of the resulting solution.

Both the bases are fully dissociated in solution.

The number of moles of OH- coming from NaOH are:

0.02 moles: 1000 ml = x moles : 300 ml

x = (6) (10-3)

The number of moles of OH- coming from Ba(OH)2,considering that each mole of the base produces two moles of OH-, are:

(2)(10-2) : 1000 = x : 200

x = (4)(10-3)

Thus the total number of moles of OH- is (6) (10-3) + (4)(10-3) = 0.01 and since the final volume of the solution is 500 ml the solution results to be 0.02 M in OH-:

pOH = - log 0,02 = 1,69

pH =14-1,69 = 12,3

6. Vinegar is a wine derivative containing acetic acid (pH = 3) and flavoring agents. Vinegar can be simulated by preparing a solution of acetic acid having pH 3 and adding appropiate flavoring agents. Calculate the ml of acetic acid ( density = 1.049 g/ml, MW = 60, Ka = 1.74 * 10-5, pKa = 4.76) needed to prepare 1 liter of vinegar.

Assuming that the concentration of the acid at equilibrium is equal to the initial concentration of the acid (Co) it has been demonstrated that:

pH = 1/2 (pKa - Log Co) = 1/2 pKa - 1/2 Log Co

and solving this equation for Log Co we have:

Log Co = (pH - 1/2pKa)*-1/2 = (3 -4.76/2)*-2

Log Co = (3 -2.38)*-2 = 0.62*-2 = -1.24

Taking the antilogarithm it results:

Co = 0.0575 M which corresponds to 0.0575 * 60 (MW) = 3.42 grams of acetic acid.

From the value of density it results that 3.42 grams corresponds to 3.45/1.049 = 3.28 ml of acetic acid .

A better result can be obtained considering that the concentration of acetic acid calculated is the concentration at equilibrium not the initial one. Considering that pH = 3 corresponds to 10-3 M H30+, the initial concentration of acetic acid is : 0.057 + 10-3 = 0.058 M thus:

ml = (0.058*60)/1.049 = 3.31

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