In: Chemistry
Identify the single most acidic or basic principle species in each of the following solutions. Calculate its molarity or number of moles (your choice). In the case of any species other than H3O+ or OH-, write the net ionic equation for the reaction of that speicies with water. Calculate the pH of each solution.
a) 10.0 mL of 0.30 M HCL
b) 10.0 mL of 0.35 M HF
c) 10.0 mL of 0.25 KOH
d) 15.0 mL of 0.30 M NaOH
e) 15.0 mL of 0.20 M NH3
a) 10.0 mL of 0.30 M HCL
no of moles of HCl = molarity * volume in L
= 0.3*0.01 = 0.003 moles
HCl --------> H+ + Cl-
0.3M 0.3M
[OH-] = Kw/[H+]
= 1*10-14/0.3 = 3.3*10-14 M
PH = -log[H+]
= -log0.3 = 0.5228
b) 10.0 mL of 0.35 M HF
no of moles of HF = 0.35* 0.01 = 0.0035moles
HF --------> H+ + F-
c) 10.0 mL of 0.25 KOH
no of moles of KOH = 0.25* 0.01 = 0.0025moles
KOH ------> K+ + OH-
0.25 M 0.25M
{H+} = Kw/[OH-]
= 1*10-14/0.25 = 4*10-14 M
PH = -log4*10-14 = 13.3979
d) 15.0 mL of 0.30 M NaOH
no of moles of NaOH = 0.3* 0.015 = 0.0045 moles
NaOH --------> Na+ + OH-
0.3 0.3
[H+] = Kw/[OH-]
= 1*10-14/0.3 = 3.33*10-14 M
PH = -log[H+}
-log3.33*10-14 = 13.4775
e) 15.0 mL of 0.20 M NH3
no of moles of NH3 = 0.2* 0.015 = 0.003 mole