a titration required 18.38mL of 0.1574 M NaOH solution. how many moles of NaOH were in this volume?
What is given is this:
18.38 mL = 0.01838 L
0.1574 M = 0.1574 mol/L
To solve for moles when given volume and molarity you do this:
Molarity * volume = moles
0.1574 mol/L * .01838 L = 0.002893 mol
It can also be written in scientific notation:
0.002893 mol = 2.893*10^(-3) mol
First, let's calculate the number of moles of NaOH used in the titration:
moles of NaOH = molarity x volume of NaOH used
moles of NaOH = 0.430 mol/L x 0.01882 L
moles of NaOH = 0.00811196 mol
Now we need to use the balanced chemical equation for the reaction between acetic acid (HAc) and sodium hydroxide (NaOH) to determine the number of moles of HAc present in the original vinegar solution:
HAc + NaOH → NaAc + H2O
From the balanced chemical equation, we can see that the mole ratio of HAc to NaOH is 1:1. Therefore, the number of moles of HAc in the vinegar solution is also 0.00811196 mol.
The molarity of acetic acid in the vinegar solution can now be calculated:
molarity of HAc = moles of HAc / volume of vinegar used
molarity of HAc = 0.00811196 mol / 0.01000 L
molarity of HAc = 0.811196 mol/L
Therefore, the molarity of acetic acid in the vinegar is 0.8112 M.
A titration required 24.62ml of 0.156m NaOH solution. How many moles of NaOH were in that volume?
To solve this problem, we'll use the same formula as in the previous question:
moles = molarity x volume
First, we need to convert the given volume from milliliters to liters:
24.62 mL = 0.02462 L
Now we can plug in the values:
moles = 0.156 mol/L x 0.02462 L
moles = 0.00384392 mol
So there were approximately 0.0038 moles (or 3.84 millimoles) of NaOH in the given volume.
Explain the function and purpose of an indicator in a titration.
An indicator is a chemical substance used in titrations to signal the endpoint of the reaction between the titrant and the analyte.
The function of an indicator in a titration is to change color when the solution being titrated has reached the equivalence point. This is important because the endpoint or equivalence point of a titration is the point at which the moles of the titrant added are stoichiometrically equal to the moles of the substance being titrated.
The purpose of using an indicator in a titration is to make it easier to detect the endpoint. The endpoint is often where the change of color occurs, signaling the reaction has come to completion.
Different indicators will change color at different pHs and each indicator has an associated pH range over which it changes color. The choice of indicator to use depends on the nature of the substance being titrated, as well as the titrant being used and the desired accuracy and precision of the results.
Overall, indicators are important to ensure accurate and precise results in a titration by signaling when the reaction has reached completion.
A sample of vinegar is titrates with a sodium hydroxide solution to find the molar it’s of acetic acid. If 18.82mL of a 0.430M NaOH solution is required to titrated a 10.00mL vinegar solution, what is the molarity of acetic acid in the vinegar?
Calculate the molarity of a KCI solution made by dissolving 21.2 g of KCI in a total volume of 500. mL.
To calculate the molarity of the KCI solution, we need to know the number of moles of KCI that was dissolved in the solution, and the volume of the solution in liters.
We can first calculate the number of moles of KCI using its molar mass:
molar mass of KCI = 39.10 g/mol (for K) + 35.45 g/mol (for Cl) = 74.55 g/mol
moles of KCI = mass of KCI / molar mass of KCI
moles of KCI = 21.2 g / 74.55 g/mol
moles of KCI = 0.2841 mol
Next, we need to convert the volume of the solution from milliliters to liters:
volume of solution = 500. mL / 1000 mL/L
volume of solution = 0.500 L
Finally, we can calculate the molarity of the KCI solution using the formula:
molarity = moles of solute / volume of solution
molarity = 0.2841 mol / 0.500 L
molarity = 0.568 M
Therefore, the molarity of the KCI solution is 0.568 M.
Choose the letter for the correct Conformation of the molecule to the right:
CH3CH2 CH, CH3
d.
a.
CH3
CH
CH3-C-CH3
b
C.
CH
CH₂
CH
CH₂
CH2-CH
CH₂
CHS
CH
e. none of these
CH₂ CH₂ CH₂