1. calculate the pH of a solution with [H3O+]=
2.4x10^-5M
2. the pOH for a KOH solution is 4.5, determine [H+] for the
solution.
3. A 15.00 mL sample of NaOH solution of unknown concentration
required 17.88 mL of 0.1053 M H2SO4 solution to reach the
equivalent point in a titratiom. what is the concentration of the
NaPH Solution?
1) pH of strong acid or stong base = -log[H+] or -log{OH-]
[H3O+] is [H+] { H+ is concentration }
pH = -log[H3O+]
pH = -log [2.4 x 10-5] = 4.62
2) KOH ----> K+ + OH-
molar ration of K+ and OH- is 1:1 so
[OH-] = [K+]
pOH = -log[OH-]
4.5 = -log[OH-]
[OH-] = 10-4.5 = 3.16 x10-5 M
[K+] = [H+] = 3.16 x10-5 M
3) Write the chemical equation for the reaction:
2NaOH + H2SO4 ---> Na2SO4 + 2H2O
The key molar ratio is that of H2SO4 to NaOH, is 1 : 2
Determine moles of H2SO4 :
moles = MV = (0.1053 mol/L) (0.01788 L) = 0.00188 mol
Determine moles of NaOH:
ratio is 1 : 2
0.00188 x 2 mol = X
X = 0.00376 mol of NaOH consumed
Determine molarity of NaOH solution:
0.00376 mol / 0.015 L = 0.25 M
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