pH is a measure of the hydrogen ion (H+) activity in a solution and, therefore, its acidity or alkalinity. Show Values for some common bases:
Solved Problems1. 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 ?
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?
remembering that M1V1 = M2V2 (10-3)(3) = (10-4)(x) 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 The number of moles of OH- coming from Ba(OH)2,considering that each mole of the base produces two moles of OH-, are:
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-:
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 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 Is NaOH acidic or basic pH?NaOH is a strong base as it completely dissolves in water to release hydroxide ions (along with sodium ions) which are responsible for the basic nature of an aqueous solution. A Compound that releases OH- ions in an aqueous solution is basic in nature. The pH value of NaOH is around 12 classifying it as a strong base.
What is pH of 1M NaOH?The pH of 1M NaOH is 13 .
What is the pH of 0.5 m NaOH?of NaOH is 0.5 , from this you can calculate pOH. Now you apply pOH+pH =14 (At 25 degree centigrade) to calculate pH of 0.5 M NaOH. So, pOH=-log[OH-] =- log(0.5) =0.3010 Hence, pH=14 - pH =13.69 .
What pH is a 10% solution of NaOH?Complete step by step answer:
pH scale ranges from 0-14, 7 is neutral, below 7 it represents acidic and above 7 it represents basic. Since, the solution is NaOH , which is basic in nature and thus it should have the value of pH more than 7 and it should be between 7 to 14. Hence, the pH of N/10 sodium hydroxide is 13.
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