be estimated. Pre-laboratory work: For the titration of 25.0 mL of 0.1 M HCI (aq
ID: 1028013 • Letter: B
Question
be estimated. Pre-laboratory work: For the titration of 25.0 mL of 0.1 M HCI (aq) with 0.1 M NaOH (aq), at what volume of NaOH (aq) should the equivalence point be reached and why? If an additional 3.0 mL of 0.1 M NaOH (aq) is then added, calculate the pH of the final solution. What is the initial pH expected for a 0.1 M solution of acetic acid? For the titration of 25.0 mL of 0.1 M acetic acid with 0.1 M NaOH (aq), at what volume of NaOH (a) is the equivalence point reached? Is the pH at the equivalence point greater than, less than or the same as in problem #1 above? Explain. 1. 2. 3. What is the initial pH expected for a 0.1 M solution of phosphoric acid (HsPO4)? 9 For 25.0 mL of 0.1 M H3Po4 (a), what volume of 0.1 M NaOH (aq) is required to fully titrate all three protons to their end points? 4. In the titration of a weak acid with a strong base, how is the half equivalence point determined and what is its significance? How are the pK and Ka of the weak acid determined from the half equivalence point? Carefully read and study the lecture text sections related to this experiment. Prepare and set up your lab notebook. Be sure to leave spaces for recording observations 5. 6. Experimental Procedure: Part 1: Strong Acid - Strong Base Titration Curve 1. Obtain about 30 mL of HCl solution and about 50 mL of NaOH solution. RecordExplanation / Answer
1)For the neutralization rxn,
HCl+NaOH ---->NaCl+H2O
mol HCl/mol NaOH=1:1
M1=0.1M
V1=25.0ml
M2=0.1M
V2=volume of NaOH
M1V1=mol of HCl
M2V2=mol of NaOH
So,M1V1=M2V2
V2=M1V1/M2= 0.1M*25ml/0.1M=25ml
or,V2=25ml =vol of NaOH at eqv point
If V2=25+3=28ml
mol of NaOH=M2V2=0.1M*28ml=(0.1 mol/L)*(0.028L)=2.8*10^-3mol
mol of HCl=M1V1=0.1M*25ml=(0.1 mol/L)*(0.025L)=2.5*10^-3mol
Excess mol NaOH=(2.8*10^-3mol) - (2.5*10^-3mol)=0.3*10^-3 mol
[NaOH]=(0.3*10^-3 mol )/(25+28)*10^-3ml=(0.3*10^-3 mol )/(25+28)*10^-3ml=(0.3*10^-3 mol )/(53)*10^-3ml=0.00566mol/L
NaOH--->OH- +Na+
pOH=-log [OH-]=-log (0.00566)=2.2
pH=14-pOH=14-2.2=11.7
pH=11.7
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.