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Author Topic: How to solve these pH problems.  (Read 25579 times)

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pokerface

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How to solve these pH problems.
« on: April 12, 2009, 03:48:03 PM »

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?
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Borek

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Re: How to solve these pH problems.
« Reply #1 on: April 12, 2009, 11:42:35 PM »

1.)What is the pH of a 0.75 M acetic acid solution? (pKa=4.74)
my answer:
HA       + OH-     ---->   H2O   + A-

Build ICE table.

Quote
2.) What is the pH of 1.50 M NH3 (aq)? (Kb=1.8 x 10^-5)
my answer: pH=11.72

Looks OK.

Quote
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?

Use Henderson-Hasselbalch equation.

Quote
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:

These are simple limiting reagent questions, don't make them harder than they are.

Quote
c.) after titrating with 20.0 mL of NaOH

You have mixed stoichiometric amounts of strong acid and strong base, you have just a solution of a neutral salt. No need for any pH calculations ;)

Quote
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).

http://www.titrations.info/acid-base-titration-curve-calculation
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