All samples of water undergo self-ionization/auto-ionization. The equilibrium fo
ID: 500147 • Letter: A
Question
All samples of water undergo self-ionization/auto-ionization. The equilibrium formed has a very small equilibrium constant, K_w = 1 times 10^-14. H_2 O (I) + H_2 O (I) H_3 O^+ (aq) + OH^- (aq) also written as: H_2 O (I) H^+ (aq) + OH^- (aq) So all aqueous solutions contain some H^+ (protons) also represented by H_3 O^+ (hydronium) and OH^- (hydroxide). The amount of each ion in a solution is related to the pH and pOH defined as: pH = -log[H^+] = -log[H_3 O^+] pH = -log[OH] Because the two quantities are related by equilibrium, it can be shown that pH + pOH = 14 for any one particular solution. Given [H^+] = 2.46 times 10^-8, calculate the pH ____ and pOH___. Acids are substances that increase the amount of H^+ in water (lowering the pH, 7). When acids and bases react, OH^- and H^+ combine to form water thus altering the pH of the resulting mixture. Strong Acids and Bases dissociate completely in water and the concentration of H^+ or OH^- is directly delated to the concentration of the add/base. 50 mL of 0.1 M HCI: pH = -log (0.1) = 1 50 mL of 0.1 M NaOH: pOH = -log(0.1) = 1 and pH = 14-1 = 13 Calculate the pH of 50 mL of 0.1 M HCI plus 10 ml 0.1 M NaOH. HCI (aq) + NaOH (aq) rightarrow H2O (I) + NaCI (aq) Calculate the pH of 50 mL of 0.1 M HCI and adding: Calculate the pH starting with 50 of 0.1 M HCI and adding: (a) 35 mL of 0.1 M NaOH (b) 50 mL of 0.1 M NaOH (c) 60mL of 0.1 M NaOHExplanation / Answer
Solution :-
a)Calculating the pH when 35 ml 0.1 M NaOH added to 50 ml 0.1 M HCl
Balanced reaction equation
HCl + NaOH ---- > NaCl + H2O
Mole ratio is 1 : 1
moles of HCl = molarity * volume in liter
= 0.1 mol per L * 0.05 L
= 0.005 mol
Moles of NaOH = 0.1 mol per L * 0.035 L = 0.0035 mol
Moles of NaOH are less so the NaOH is limiting therefore moles of HCl remain after reaction are calculated as
Moles of HCl remain after reaction = 0.005 mol – 0.0035 mol = 0.0015 mol HCl
New molarity of HCl = 0.0015 mol / (0.035 L + 0.05 L) = 0.01765 M
pH= -log [H+]
pH= -log [0.01765]
pH= 1.75
b) calculating the pH when 50.0 ml 0.1 M NaOH is added
moles of HCl = molarity * volume in liter
= 0.1 mol per L * 0.05 L
= 0.005 mol
Moles of NaOH = 0.1 mol per L * 0.05 L = 0.005 mol
Both acid and base have same moles
So they completely neutralize each other
So the pH of the neutral solution = 7.00
Part c) calculating the pH when 60 ml of 0.1 M NaOH added to 50 ml 0.1 M HCl
moles of HCl = molarity * volume in liter
= 0.1 mol per L * 0.05 L
= 0.005 mol
Moles of NaOH = 0.1 mol per L * 0.06 L = 0.006 mol
Here moles of NaOH are more than moles of HCl
So the moles of NaOH remain after reaction = 0.006 mol – 0.005 mol = 0.001 mol NaOH
New molarity of NaOH = 0.001 mol / (0.06 L+0.05 L) = 0.00909 M
pOH = -log [OH-]
pOH = -log [0.00909]
pOH = 2.04
pH = 14 – pOH
pH= 14 – 2.04
pH= 11.96