Calculate pH? Lookin for the best answer!?
Calculate pH:
25mL of .2 M CH3OOH + 10 mL of .2 M NaOH? Ka= 1.8E-5
Step-by-step procedure please! I have to do 20 of these!
Moles of acid:
0.025L * 0.2mole/L = 0.005 moles
Moles of NaOH:
0.01L * 0.2mole/L = 0.002moles
Concentration of acid:
0.005mol/0.030L = 0.17M
Concentration of NaOH:
0.002mol/0.030L = 0.07
Ka= [CH3COO-][H+]/ [CH3OOH]
1.8E-5 = X^2/0.17
X= 0.00175M = [H+]
*NaOH dissociates completely there for [NaOH] = [Na+] = [OH-] = 0.07M
In solution:
[OH-] - [H+] = 0.068[OH-]
pOH = -log(0.068) = 1.17
pH = 14 - pOH = 12.83