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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 mlx = (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)*-2Log 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