KOH + HBr ---> KBr + H2O
0,3 moles of HBr ---in-------1000ml
x moles of HBr-------in------75ml
x = 0,0225 moles of HBr
according to the reaction: 1 mole of KOH = 1 mole of HBr
so
0,0225 moles of HBr = 0,0225 moles of KOH
0,0225 mole of KOH------in-----45ml
x moles of KOH -----------in------1000ml
x = 0,5 moles of KOH
answer: 0,5 mol/dm³ KOH (molarity)
<u>Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM</u>
Given:
Acetic Acid/Sodium Acetate buffer of pH = 5.0
Let HA = acetic acid
A- = sodium acetate
Total concentration [HA] + [A-] = 250 mM ------(1)
pKa(acetic acid) = 4.75
Based on Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77
[A-] = 1.77[HA] -----(2)
From (1) and (2)
[HA] + 1.77[HA] = 250 mM
[HA] = 250/2.77 = 90.25 mM
[A-] = 1.77(90.25) = 159.74 mM
Question 1:
(a) Sulfurous acid: H2SO3
Sulfuric acid: H2SO4
(b) Nitrous acid: H2NO2
Nitric acid: H2NO3
Question 2:
To calculate the pH, based on concentration of H+ ions, there is one formula:
So the pH of this solution is
(the solution is basic).
