I have some of these worked out,but is unsure if it's right or not. Please check them. But there were some that I have no idea. Please help. THank you! I really appreciate it:D
1.)What is the pH of a 0.75 M acetic acid solution? (pKa=4.74)
my answer:
HA + OH- ----> H2O + A-
0.75 M
pH= pKa +log(base/acid)
= 4.74 + log (
2.) What is the pH of 1.50 M NH3 (aq)? (Kb=1.8 x 10^-5)
my answer: pH=11.72
NH3 (aq) + H2O (aq) --> NH 4+ (aq) + OH-(aq)
1.50 M ---- 0 0
-x +x +x
-------------------------------------------------------------
1.50-x +x +x
1.50-.005196= 1.555 .005196 .005196
x2/(1.50-x)=1.8 x 10^-5
x2= (1.8x10^-5)..x= .005196
pOH= -log[OH-]
= -log[.005196]
pOH= 2.284
pOH + pH = 14
pH= 11.72
3.) a buffer solution is 0.35 M NH3 and 0.45 M NH4Cl (Kb for NH3=1.8 x 10^-5). What is the pH of this buffer?
b.) If 0.0050 mols HCl is added to 0.500 L of this buffer solution, what will be the pH?
4.) calculate the pH at the following points in the titration of 40.00 mL of 0.50 M HCl with 1.00 M NaOH:
a.) Before the addition of any NaOH (Initial pH)
HCl (aq) + H2O ----> Cl- (aq) + H3O+(aq) MhclVhcl=MnaohVnaoh
.50M ----
b.) after titrating with 10.0 mL of NaOH
HCl + NaOH --> NaCl + H2O HCl: (.50 M) (.040 L)= 0.02 mol
.02 mol 0.01 mol --- NaOH: (1.00 M)(0.010L)= 0.01 mol
-0.01 mol -0.01 mol 0.01 mol [HCl]= mol/V, 0.01 mol/0.050 L= .2 M
----------------------------------------------------------------
0.01 mol 0 0.01mol
HCl + H2O ---> H3O+ + Cl-
.2M .2 M .2 M
------------------------------------------------------------
pH= -log[H3O+]
= -log[.2M]
pH= .699
c.) after titrating with 20.0 mL of NaOH
HCl + NaOH ---> NaCl + H2O
0.02mol 0.02 mol NaOH= (1.00M)(.020L)=0.02 mol
-0.02mol [NaOH]= mol/V, 0.02mol/0.060L= .333 M
------------------------------------------
0mol 0.02 mol
HCl + H2O ---> H3O+ + Cl-
.
------------------------------------------------------------
[H3O+]=kw/[OH-]
=1 x10^-14/.333M
[H3O+]= 2.00 x 10^-14 M
pH= -log[H3O+]
= -log[2.00 x 10^-14 M]
pH= 13.70 i think tjhos is wrong:(
d.) after titrating with 30.0 mL of NaOH
HCl + NaOH ---> NaCl + H2O
0.02mol 0.03 mol NaOH= (1.00M)(.030L)=0.03 mol
-0.02mol -0.02mol
------------------------------------------
0.01mol
HCl + H2O ---> H3O+ + Cl-
.
------------------------------------------------------------
[H3O+]=kw/[OH-] [NaOH]= mol/V, 0.01mol/0.070L= .143 M
((
5.) Calculate the pH at the following points in the titration of 50.00 mL of 0.750 M CH3COOH with 0.500 M NaOH: (Ka for acetic acid is 1.8 x 10^-5).
a.) before the addition of any NaOH (Initial pH)
HC2H3O2 + H2O ---> C2H3O2 + H3O+
b.) after the addition of 25.00 mL of 0.500 M NaOH
c.) after the addtion of 37.50 mL of 0.500 M NaOH
d.) what volume of NaOH is needed to reach equivalence point?
M1V1=M2V2
0.050L*0.750 M=0.500MV2
V2= 0.075 L
e.) what is the pH at the equivalence point?
Kb =Kw/Ka
1x10^-14/1.8 x10^-5= 5.56 x 10^-10
[OH-]=
pOH= -log [OH-]
f.) what is the pH after adding 125.00 mL of NaOH?