mol NaOH= 31.43mL x 1L/1000mL x 0.108 mol/L = 0.00339 mol NaOH
*There is a 1:1 Ratio between AC (Acetic Acid) and NaOH therefore,
mol AC = 0.00339 mol NaOH x 1 mol AC/1 mol NaOH = 0.00339 mol AC
Mass of AC in Vinegar = 0.00339 mol AC x 60.05 g AC/1 mol AC = 0.204 g AC
% AC in Vinegar = (g AC/g Vinegar) = 0.204 g AC/4.441 g Vinegar) x 100% = 4.6%
Andrea
Palm Beach State College
1. The problem statement, all variables and given/known data
The problem is as follows:
100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide . An appropriate indicator is used. Ka for acetic acid is 1.7 x 10 -5. Calculate the pH in the flask at the following points in the titration.
a. when no NaOH has been added.
b. after 25.0 ml of NaOH is added
c. after 50.0 ml of NaOH is added
d. after 75.0 ml of NaOH is added
e. after 100.0 ml of NaOH is added
2. Relevant equations
According to my text:
(Liters of acetic acid solution)X(mol H+/1Liter of solution)=mol H+
(Liters of sodium hydroxide solution)X(mol OH-/1Liter of solution)=mol OH-
(Liters of acetic acid)+(Liters of hydroxide solution)=total Liters
Concentration of H+=(moles H+/total Liters)
pH=-log(concentration of H+ *above*)
3. The attempt at a solution
What I have so far is this, but my pH answers are nowhere near what they should be. What am I missing?
"100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide . An appropriate indicator is used. Ka for acetic acid is 1.7 x 10 -5Calculate the pH in the flask at the following points in the titration."
(0.1L soln)(0.100M H+/1L soln)=1.000 X 10-2
a. "when no NaOH has been added" pH=-log(0.100)=1.000
b. "after 25.0 ml of NaOH is added" (0.025 L soln)(0.100M OH-/1L soln)=2.5 X 10-3 mol OH-
100.0mL+25.0mL=125.0mL=0.125L [H+]=(7.5x10-3)/(0.125L)=6.000X10-2M pH=-log(6.000X10-2M)=1.222
c. "after 50.0 ml of NaOH is added" (0.050L soln)(0.100M H+/1L soln)=5.000 X 10-3 mol OH-
100.0mL+50.0mL=150.0mL=0.150mL [H+]=(5.000 X 10-3)/(0.150L)=3.333 X 10-2 pH=-log(3.333 X 10-2)=1.477
d. "after 75.0 ml of NaOH is added" (0.075 L soln)(0.100M H+/1L soln)=7.5X10-3 mol OH-
100.0mL+75.0mL=175.0mL=0.175L [H+]=(2.5 X 10-3)/(0.175L)=1.429X10-2 pH=-log(1.429X10-2)=1.845
e. "after 100.0 ml of NaOH is added" (0.1L soln)(0.100M H+/1L soln)=1.000 X 10-2 mol OH-
100.0mL+100.0mL=200.0mL=0.200L [H+]=pH=7.0
This is a table similar to that which is shown in my text, where I have filled in my answers.
H+(aq) + OH-(aq) --> H2O(l)
Before addition 1.000 X 10-2 0 ---------
Addition 25.0 ml 2.5 X 10-3
Addition 50.0 ml 5.000 X 10-3
Addition 75.0 ml 7.5X10-3
Addition100.0 ml 1.000 X 10-2
After Adn25.0 ml 7.5x10-3 0 --------
After Adn 50.0 ml 5.000 X 10-3 0 --------
After Adn 75.0 ml 2.5 X 10-3 0 --------
After And 100.0 ml 0 0 ---------
ollololo
0.0575
HCL is used to provide the acidic conditions required in the iodometeric titration.
need of pilot reading in volumetric analysis
H2SO4 is added in reaction because acidic medium is required and it provides proton to the reaction .
Titration is the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration. Titration provides the equivalent volumes of acidic and basic solutions. In order to find this, MaVa/Ca formula needs to be used. In titration, when equal numbers of H3O+ and OH- from the acidic and basic solutions react, the resulting solution is neutral (water and salt). In titration, the end point would be the point at which the indicators change color; in this case the indicator turned pink. The equivalence point would the point at which the two solutions used in titration are present in chemically equivalent amounts. The indicator, phenothaylene, is used to determine the equivalence point of weak-acid/strong- base titrations.
endpoint
calculate the amount of NaOH required in neutralising the sulphonated mass. calculate the remaining sulphuric acid stoichiometrically. from that you get percent sulphonations. or take sample and carry out titration
HCL is used to provide the acidic conditions required in the iodometeric titration.
need of pilot reading in volumetric analysis
How do you calculate the production line personnel required?
Draw a straight line and place zero degrees of the protractor at the endpoint of the line then mark out 135 degrees and join the endpoint with the marked out degrees to construct the required angle.
H2SO4 is added in reaction because acidic medium is required and it provides proton to the reaction .
Endpoint security, also known as endpoint protection, prevents malware and other threats from reaching their destination by scanning a computer's hard drive or running application before they reach it. Endpoint protection has two significant benefits: the need to install protection software on individual devices is no longer required (unless perhaps devices are being transferred from home to office) and device management becomes drastically simplified. Endpoint security services can be divided into five different functional areas: Host-Based Protection Application Control Data Aggregation Protection Threat Hygiene and Management Cloud Connectivity
Its very easy please follow the steps below, 1. Calculate the DN Required by using Hydrauic calculation 2. Calculate the PN required 3. Calculate the SN required 4. Use the mixuture formula to decide the composition 5. Convert this into Kg/m 6. Use the classical laminate theory to calculate the thicness by forming a nine by nine matrix
By calculate do you mean calculate the connections required or phasor diagrams? full question would be helpful
The length of pipe is required to be known to help procure the required quantity of pipe.
Titration is the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration. Titration provides the equivalent volumes of acidic and basic solutions. In order to find this, MaVa/Ca formula needs to be used. In titration, when equal numbers of H3O+ and OH- from the acidic and basic solutions react, the resulting solution is neutral (water and salt). In titration, the end point would be the point at which the indicators change color; in this case the indicator turned pink. The equivalence point would the point at which the two solutions used in titration are present in chemically equivalent amounts. The indicator, phenothaylene, is used to determine the equivalence point of weak-acid/strong- base titrations.
The initial reading on the buret is 10.45mL and the final reading is 40.55mL. Subtract final from initial your answer. 40.55mL-10.45mL=30.10mL