Can you please help me make sure the Molarity of NaOH based on my answers from t
ID: 859175 • Letter: C
Question
Can you please help me make sure the Molarity of NaOH based on my answers from the final-initial=volume is correct? Then How do I calculate the NaOH molarity for one trial?
Can you please help me make sure the Molarity of NaOH based on my answers from the final-initial=volume is correct? Then How do I calculate the NaOH molarity for one trial? Next, can you show me how to calculate the molar concentration of vinegar? How do I calculate the molarity of vinegar for one trial? Preparation of NaOH solution: Mass of sodium hydroxide NaOH expected to be needed to make 250mL of 0.3 M NaOH solution Calculation: Standardization of NaOH solution: Initial volume of NaOH Final volume of NaOH Volume of NaOH used Molarity of NaOH Average concentration of NaOH Concentration of HCl Calculation of NaOH molarity for one Trial: Titration of Household Vinegar Initial volume of NaOH Trial 1 Trial 2 Trial 3 Final volume of NaOH Volume of NaOH used Molar concentration of vinegar Average molar concentration of vinegar Calculation of molarity of vinegar for one Trial:Explanation / Answer
hi,
its simple to calculate molarityof NaOH used to neutralise acid.
in 1000mL NaOH there are 0.3 moles.
so for trail 1 you have used 13.44 ml
so molar concentration used to neutrilise acid is = (13.44*0.3)/1000 = 0.004032 mole of NaOH
to find the HCl concentration; take average of 3 molar values.
0.004032 mole of NaOH = 0.004032 mole of HCl
0.004032 mole of HCl in XXXX ml of solution (XXXX is the ml you took in conical flask, generally 20 ml)
moles of HCl present in 1000mL are (0.004032 x 1000) / XXXX = yyyy M HCl.
in standardisation you have to compare Molar value of HCl obtained to the molar value of HCl given on experiment. it should be some what equal.
To find molar concentration of Vinegar:
for trial 1:
volume used to neutrilise vinegar = 12.75 -3.50ml
= 9.25ml
molar concentration used to neutrilise vinegar is = (9.25*0.3)/1000 = 0.002775 mole of NaOH
0.002775 mole of vinegar in XXXX ml of solution (XXXX is the ml you took in conical flask, generally 20 ml)
moles of vinegar present in 1000mL are (0.002775 x 1000) / XXXX = yyyy M vinegar .
finally take the average.
in genral for standardisation the Molar vale of HCl will be given to you. in that case use the below method.
theoritically you can reverse if you know the HCl concentration to to find the NaOH moles like the one shown below.
in 20 ml the HCl present is (20*0.132)/1000 = 0.00264M
so 0.00264M HCl = 0.00264M NaOH.
moles of NaOH present in 1000mL are ( 0.00264 x 1000) / 13.44= 0.19M of NaOH.