To create a theoretical titration curve for the neutralization of a strong acid
ID: 527964 • Letter: T
Question
To create a theoretical titration curve for the neutralization of a strong acid by a strong base. 1) Complete the Table shown below by calculating the missing information. (Be sure to provide example calculations that illustrate clear understanding of the mathematical operations applied to the Table.) Create a plot by graphing pH (y-axis) versus Volume (mL)_NaOH (x-axis). Use graph paper to hand draw the graph or use Microsoft Excel to create plot and attach printout of plot. (Attach Graph) Then, create a second plot by graphing Delta pH/Delta Volume_NaOH (y-axis) versus mL NaOH_added (x-axis). This rate change plot is an alternative means to identify the Equivalent Point for the titration. (Attach Graph) Titration of 40.0 mL of 0.100M HCl (strong acid) with 0.100M NaOH (strong base). Example calculations (use additional pages, if necessary):Explanation / Answer
when 0.0 ml NaOH is added then total HCl present is 40 ml and strength is 0.1 M. So mmol of HCl=40*0.1=4
[H3O+]=4/1000=0.004 So PH=-log[H+]=-log(4*10^-3)=3-log4=3-2log2=3-2*0.3010=3.0-0.6020=2.3980
POH=14-2.3980=11.602 Thus [OH-]=10^-11.602
when 10 ml o.1 M NaOH is added mmol of NaOH added=10*0.1=1 So mmol of HCl left=4-1=3
[H3O+] after neutralisation=0.003 So PH=-log(3*10^-3)=3-log3=3-0.477=2.523 So POH=14-2.523=141.477
[OH-]=10^-11.477=3.33*10^-12
When 20 ml NaOH is added mmol of NaOH added=20*0.1=2 mmol of HCl left=2 {H3O+]=2/1000=0.002 So PH=-log(2*10^-3)=3-log2=3-0.3010=2.70 POH=14-2.70=11.30 [OH-]=10^-11.30=5.011*10^-12
When 30 ml NaOH added mmol of NaOH=30*0.1=3 HCl left=4-3=1 [H3O+]=1/1000=0.001 PH=-log(1*10^-3)=3
So POH=14-3=11 [OH-]=10^-11
When 39 ml NaOH added mmol NaOH added=39*0.1=3.9 Left HCl=4-3.9 mmol=0.1 mmol [H+]=0.1/1000=10^-4
PH=-log(10^-4)=4 POH=14-4=10 [OH-]=10^-10
When 39.9 ml NaOH added mmol NaOH added=39.9*0.1=3.99 mmol HCl left=4-3.99=0.01 [HCl]=0.01/1000=10-5 PH=-log(10^-5)=5 POH=14-5=9 [OH-]=10^-9
When 40 ml NaOH added then mmol of NaOH added=40*0.1=4 Left acid =4-4 mmol=0 mmol PH=POH=7 [H+]=[OH-]=10^-7
When 50 ml NaOH added mmol of base=50*0.1=5 excess base after neutralising 4 mmol acid=5-4=1 [OH-]=1/1000=10^-3 POH=3 PH=14-3=11 [H+]=10^-11
Similarly all can be calculated.
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