Questions LLC
Login
or
Sign Up
Ask a New Question
Chemistry
Laboratory Techniques
Titration
can u help me in explaining why we used 5ml instead of 55.00ml of acetic acid on our calculation in titration ?
1 answer
I could if you supplied enough information; with what you've supplied it's a shot in the dark. I have no ida what you did.
You can
ask a new question
or
answer this question
.
Similar Questions
Calculate the amount (in mL) of 1.520M NaOH that is required to add the following acetic acid solution to prepare a buffer with
Top answer:
You had this in math. It isn't chemistry. 5.75 = 4.74 + log b/a 1.01 = log b/a 10^(b/a) = 1.01 Plug
Read more.
Calculate the amount (in mL) of 1.520M NaOH that is required to add the following acetic acid solution to prepare a buffer with
Top answer:
I would do this. 5% w/v means 5g HAc/100 mL solution which is (5/60) mols/100 and that is 5/60/0.1L
Read more.
Tf (acetic acid) = 16.1°C
Tf (with 3.00g of solute in 10.00mL of acetic acid) = - 2.5°C Use the data to calculate the molecular
Top answer:
OK, you have molality = m m = mol/kg solvent. You have m and kg solvent, substitute and solve for
Read more.
What is the molarity of a solution of acetic acid if 35.00mL is titrated to the end point with 68.20ml of 0.750M KOH? What is
Top answer:
acetic acid is HAc HAc + KOH ==> KAc + H2O mols KOH = M x L Look at the coefficients. mols HAc =
Read more.
What is the pH of the resulting solution if 30.00 mL of 0.100M acetic acid is added to 10.00mL of 0.100 M NaOH? For acetic acid,
Top answer:
The equation is CH3COOH + NaOH ==> CH3COONa + H2O (What you wrote--C6H5COOH-- is benzoic acid). I
Read more.
What is the pH obtained by mixing 50.00mL of 0.100N HOAc (acetic acid) and 25.00mL of 0.200N NaOH?
Top answer:
Use the Henderson-Hasselbalch equation.
Read more.
A 10.0 ml sample of vinegar which is an aqueous solution of acetic acid, requires 16.5ml of a 0.500M NaOH solution to reach the
Top answer:
.303m
Read more.
Titration of 50.0 mL of acetic acid reaches equivalence after delivery of 22.5mL of standardized NaOH 0.21 M.
What is the initial
Top answer:
HAc + NaOH ==> NaAc + H2O First, determine the molarity of the HAc. mL x M = mL x M 22.5 x 0.21 =
Read more.
calculate the concentration of an acetic acid solution prepared by mixing 13.5ml of 10.0M acetic acid with 250.0ml
of 0.15M NaCl
Top answer:
10.0 M x [(13.5 mL)/(13.5 + 250.0)] = 10.0 x (13.5/263.5) = ? M
Read more.
Calculate the concentration of an acetic acid solution prepared by mixing 13.5mL of 10.0M acetic acid with 250.0mL of .15M NaCl
Top answer:
I don't see how the NaCl will affect the M except by dilution. 10.0 M x (13.5/(250+13.5) = ?
Read more.
Related Questions
What is the pH of a 0.25 M acetic acid solution if the Ka of acetic acid is 1.8 x 10-5?
2.67 0.60 5.34 11.33 I got A, 2.67. I
Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25°C. Calculate the molarity of a
Acetic acid is a weak acid with the formula , CH3COOH, the Ka for acetic acid is 1.76 x 10-5.
In aqueous solution, acetic acid
You're performing a titration of some acetic acid. You have reason to doubt that the grocery store is selling you vinegar at the
Lab: Determining Ka of Acetic Acid
Purpose: The purpose of this experiment is to determine the molar concentration of a sample of
calculate the molarity of acetic acid in a vinegar sample,knowing that 5.00ml of vinegar requires 43.50ml of 0.105 M NaOH to
Acetic acid (HC2H3O2) is an important component of vinegar. A 10.00mL sample of vinegar is titrated with .5052 M NaOH, and 16.88
In Chemistry: what is ph during the titration of 20.00ml of 0.1000m triethylamine (ch3ch2)3n (kb=5.2x 10-4) with 0.1000m HCL
Hydrochloric acid is a strong acid. Acetic acid is a weak acid. Which statement about hydrochloric acid and acetic acid is
Hydrochloric acid is a strong acid. Acetic acid is a weak acid. Which statement about hydrochloric acid and acetic acid is