Question

1. IN A 0376 EXP. 20.0 mL OF .400M HA REACTS WITH 10.0 mL OF .200M OH^- AND DILUTED TO 200.0 mL . WHAT IS THE INITIAL ROW OF THE STOICH. TABLE: HA + OH^- --> A^- IN ORDER

   A).0400,.0100,0 B) .0400,.0100,.0100 C) .400,.200,0 D) .00800,.00200,0

2. IN A 0376 EXP. THE FIRST LINE IN THE STOICH. TABLE FOR A GIVEN RUN IS: HA=.100M OH=0.0600M A=0M WHAT IS [ A^-] ? HA + OH^- --> A^-

A).0400M B) 0 C) .0600M D) .160M

3. IN A 0376 EXP. THE MIXTURE HAS A pH= 3.00. WHAT IS [ H⁺ ] ?

   A)3.0X10³M B) 1.0X10¹¹M C) 3.00M D) 1.0X10^-3M

4. IN A 0376 EXP. THE MIXTURE HAS [ A^- ] = .0250M. IT WILL BE PLOTTED ON THE Y-AXIS AS

   A).0250 B) 40.0 C) 15.8 D) -1.60

5. IN A PLOT IN THE 0376 EXP. THE SLOPE=2.0X10⁵ AND (HA)_I = .500M. WHAT IS THE pK_A ?

   A)5.00 B) 1.0X10^-5 C) 5.30 D) 5.60

Answer 1

(1) Option D is the correct answer.

For HA, concentration=0.400M/L. Volume=20ml=0.02L

Therefore, the number of moles of HA, n=Concentration×Volume= 0.400×20/1000= 0.008

For OH-, concentration=0.200M/L

Volume=10ml=10/1000=0.01L

Hence , the number of moles of OH- = CV= 0.200×10/1000=0.002

After dilution , the number of moles of A- will be zero.

(2) Option B is the right answer. Since the stochiometry of A- is 0, its concentration will be zero. (Concentration= number of moles/Volume)

(3) Option D is the correct answer.

We know that, PH= -log[H+]

Here PH=3

So, 3= -log[H+]

Log[H+]= -3

[H+] =antilog [-3]=1.0×10^-3

(4) Option D is the right answer. On Y axis we will take log[A-]=log[0.0250]= -1.60

(5) Option B is the right answer.

Slope of the plot= PKa/[HA]

PKa= Slope×[HA]= 2×10^-5× 0.500= 1×10^-5