Acid-Base Shtuff

Lester

100

Is this solution a buffer? Why or why not?

50.0 ml of 0.100 M HNO₃ and 50.0 ml of 0.500 M NaC₂H₃O₂

R HNO₃    +   NaC₂H₃O₂     --->    NaNO₃ + HC₂H₃O₂

I  5 mmol     25 mmol                 0             0

C -5             -5                          +5           +5

E  ~0            20                         5              5

Solution contains a ratio of acid/conjugate that is greater than 0 and less than 100. Solution is a buffer

200

300.0 ml of NaC₂H₃O₂ (aq) is mixed with 700.0 ml of 0.300 Hcl. Calculate the pH

R  NaC₂H₃O₂     +   HCl  -->     +   HC₂H₃O₂ +  NaCl

I   150mmol         210mmol           0                  0

C  -150                 -150                +150          +150

E  ~0                    60                   150               150

pH = -log (60mmol/100m) = 1.22

300

Calculate the pH of the solution after the following are mixed: 300.0 ml of 0.200M KOH and 200.0 ml of 0.400M HC₂H₃O₂ (Ka = 1.8 x 10^-5)

R   HC₂H₃O₂ + KOH   ---> H₂O + KC₂H₃O₂

I    80.0 mmol  60.0mmol   lots       0

C    -60.0          -60.0        +60.0     +60.0

E    20.0            ~0            +60        +60

Acid-base conjugate pair present

pH = pKa + log(conj base/acid)

pH = -log(1.8x10^-5) + log(0.12/0.04)

pH = 5.22

Harvey

100

Determine the pH of a solution 0.0025M solution of NH₃ (aq) (Kb = 1.8 x 10^-5)

R      NH₃       +        H₂O        --->     NH₄⁺      OH^-

I       0.0025M       lots                       0             0

C      -x                                            +x          +x 

E      0.0025-x                                    x             x

1.8 x 10^-5 = x²/0.0025-x

x = 2.12 x 10^-4

pOH = -log(2.12 x 10^-4) = 3.67

pH = 10.33

200

Calculate the pH of a solution that is 0.250M KCN

Ka = 2.1 x 10^-9

R   CN^-      +        H₂O     --->     HCN       +      OH^-

I    0.250             Lots                0                    ~0

C    -x                  -x                  +x                  +x

E    0.250-x                               x                      x

4.76 x 10^-6 = x²/(0.250-x)

x = 6.9 x 10^-4

pOH = -log (6.9 x 10^-4) = 3.16

pH = 14.00 - 3.16 = 10.84

300

Mix 75.0 ml of 0.5M HC₂H₃O₂ and 25.0 ml of 1.00M KOH. Calculate the pH.

Ka = 1.8 x 10^-5 for acetic acid

R  HC₂H₃O₂   +   KOH   --->   KC₂H₃O₂     +    H₂O

I   37.5 mmol     25.0 mmol      0                 lots

C   -25.0            -25.0            +25              +25

E    12.5               0                  25                 lots

It's buffer time!

pH = pKa + -log([A-]/[HA])

pH = 5.05

Steeeeeeve

100

Classify the following as strong acids (SA), weak acids (WA), strong bases (SB), or weak bases (WB)

HI

KOH

HF

NH₃

(CH₃)₃NH⁺

SA

SB

WA

WB

WA

200

Determine if the solution is acidic, basic, neutral, or cannot be determined from the information given. (Unless otherwise indicated, all solutions are 25ºC). 

1. pH = 3.8

2. pH > 4.00

3. [H+] = 3.6 x 10^-10

4. 2.00 x 10^-9 M HCl

5. At a temperature other than 25ºC, a solution has a higher [OH^-] than [H⁺] and a pH of 7

6. 40.0 ml of 1.5 M KOH is mixed with 60.0 ml of 1.0 M HF (Ka = 6.8 x 10^-4)

7. pH = 6.90 at 37ºC (human body temperature; Kw = 2.5 x 10^-14)

1. A

2. Not enough information

3. B

4. A, almost neutral

5. B

6. B

7. B, take the pKw/2 to determine the neutral pH

300

Mix 100.0 ml of 0.500M HC₂H₃O₂ and 100.0 ml 0.250 M KOH

Ka = 1.8 x 10^-5

What is the pH?

R HC₂H₃O₂     +    KOH   -->    H₂O      +     KC₂H₃O₂

I  50.0 mmol       25mmol                             0

C -25                  -25              +25              +25

E  25                  ~0                                    25

pH = pKa + log([A^-]/[HA])

pH = 4.74