Question as follows:

Question

Question as follows:

75.0g tartaric acid in 1 Liter is what molarity(moles/liter) mw=150.087 g/mol
M1=
25 ml tartaric solution is titrated with 31 ml ammonium.
What is the equation rearanged for normality of ammonium: starting with N1V1=N2V2
N1=
V1=
V2=
Final Answer N2=

THANKS! I can't wrap my head around this!

Answer 1

M = mols/L and mols = g/molar mass.

Therefore, mols H2T(tartaric acid) = 75.0/150.087 = ?
and M = mols/L.

Nomality of H2T = 2*M
Then 25*N H2T = 31*N NH3. Solve for N NH3.
By the way I assume this is ammonia, the ammonium ion doesn't exist by itself.

Answer 2

Thank you DrBob222!

Answer 3

To find the molarity of the tartaric acid solution, we can use the formula:

Molarity (M) = moles of solute / volume of solution in liters

Given:
Mass of tartaric acid = 75.0 g
Molar mass of tartaric acid = 150.087 g/mol
Volume of solution = 1 L

First, we need to find the moles of tartaric acid using the formula:

moles = mass / molar mass

moles of tartaric acid = 75.0 g / 150.087 g/mol

moles of tartaric acid = 0.49971 mol

Now we can calculate the molarity:

Molarity = 0.49971 mol / 1 L

Molarity = 0.49971 M (moles/liter)

Now, let's move on to the titration part of the question. Here's the rearranged equation N1V1 = N2V2 for normality calculations:

N1 * V1 = N2 * V2

We are given:
V1 = 25 mL
V2 = 31 mL

Now let's calculate N1:

N1 * 25 mL = N2 * 31 mL

Now, divide both sides by 25 mL:

N1 = (N2 * 31 mL) / 25 mL

Simplifying further:

N1 = (31/25) * N2

N1 = 1.24 * N2

So, the equation rearranged for the normality of ammonium is: N1 = 1.24 * N2

To find the final answer for N2, we need more information. The molarity or normality of the ammonium solution is needed. If you provide that information, we can calculate the final answer for N2.