Chemistry 106 83 Plotting the inverse of the conjugate acid concentration vs hyd

Question

Chemistry 106 83 Plotting the inverse of the conjugate acid concentration vs hydronium concentration (remember [H H30 ]) we can obtain the slope for the straight line equation: Slope Ka HAO From the graph and slope value, knowing the acid concentration before neutralization we can determine the Ka. 1. Using the above information and the data given below, calculate all the required quantities in the table 2, Calculate the average Ka. 3. Graph, obtain the slope and calculate the Ka value for the weak acid. 4. Identify the acid based on the dissociation constants you have found. Use the table in the back of your lab manual. 5. Compare the Ka values obtained from the two methods (graphing method and calculation method). Data A student is trying to identify a weak acid by measuring the pH of partially neutralized weak acid solutions. In the lab the student is preparing three solutions by adding 5.00 mL, 10.00 mL, and 15.00 mL of a 0.200 M NaOH solution to 10.00 mL of 0.500 M weak acid solution. The solutions are labeled A, B and C below. The student diluted the solutions to 100.0 mL using the deionized water and measured the pH for each solution. The pH values are given in the table below. Solution pH [H,0l ml. 0.200 M NaOH [A] e HA eq1/IA] K. 3.21 3.66 5.00 10.00 15.00 4.01 You must include all data you collected or you analyzed, all your calculations, all tables, graphs or charts from Parts 1 and 2 in your report. Experiment 11

Explanation / Answer

Let the weak acid be HA,

HA(aq)+H2O(l) <--->A-(aq)+H3O+(aq)

pH=-log [H3O+]

So,[H3O+]=10^-pH ..........(1)

Ka=[A-]eq [H3O+]eq/[HA]eq .......(2)

But [A-]eq]=[H3O+]eq ........(3)

[HA]eq=[HA]o-[H3O+]eq .......(4)

Using eqn (1) to (4) , the data table values can be calculated out,

Solution A) (5.00ml of 0.2M NaOH )+ (10ml of 0.5M HA)

mol of NaOH=0.005L*0.2mol/L=0.001 mol

mol HA=0.010L*0.5mol/L=0.005 mol

So, NaOH will neutralize 0.001mol HA as per the equation: NaOH+HA--->NaA+H2O

remaining HA=0.005-0.001=0.004mol

mol A- formed=mol of NaOH reacted=0.001 mol

[A-]=0.001/0.015L=0.0667M

[HA]o=0.004mol/total volume=0.004mol/15ml=0.004mol/0.015L=0.267M

Thus, [HA]eq=[HA]o -[H3O+]

[H3O+]=10^-pH=10^-3.21=0.000617 M (for diluted sample )

[A-]eq=[H3O+]eq=0.000617 M

[HA]eq=0.267M-0.000617 M=0.266M

1/[A-]=1/(0.0667MM)=15.0 M^-1

Ka=(0.000617 M)(0.000617 M)/(0.266M)=1.431*10^-6

Solution B)

(10.00ml of 0.2M NaOH )+ (10ml of 0.5M HA)

mol of NaOH=0.010L*0.2mol/L=0.002 mol

mol HA=0.010L*0.5mol/L=0.005 mol

So, NaOH will neutralize 0.002mol HA as per the equation: NaOH+HA--->NaA+H2O

remaining HA=0.005-0.002=0.003mol

mol A- formed=mol of NaOH reacted=0.003 mol

[A-]=0.002mol/0.020L=0.1M

[HA]o=0.003mol/total volume=0.003mol/20ml=0.003mol/0.020L=0.15M

Thus, [HA]eq=[HA]o -[H3O+]

[H3O+]=10^-pH=10^-3.66=0.000218 M (for diluted sample )

[A-]eq=[H3O+]eq=0.000218 M

[HA]eq=0.15M-0.000218 M=0.1498M

1/[A-]=1/(0.1M)=10.0 M^-1

Ka=(0.000218M)(0.000218M)/(0.1498M)=3.172*10^-7

solution C)

(15.00ml of 0.2M NaOH )+ (10ml of 0.5M HA)

mol of NaOH=0.015L*0.2mol/L=0.003 mol

mol HA=0.010L*0.5mol/L=0.005 mol

So, NaOH will neutralize 0.003mol HA as per the equation: NaOH+HA--->NaA+H2O

remaining HA=0.005-0.003=0.002mol

mol A- formed=mol of NaOH reacted=0.003 mol

[A-]=0.003mol/0.025L=0.12M

[HA]o=0.002mol/total volume=0.002mol/25ml=0.002mol/0.025L=0.08M

Thus, [HA]eq=[HA]o -[H3O+]

[H3O+]=10^-pH=10^-4.01=0.0000977 M (for diluted sample )

[A-]eq=[H3O+]eq=0.0000977M

[HA]eq=0.08M-0.0000977 M=0.0799 M

1/[A-]=1/(0.12M)=8.333M^-1

Ka=(0.000097M)(0.000097M)/(0.0799M)=1.177*10^-7

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