I kept telling you there had to be some connection somewhere. Tartaric acid is H2T.
H2T + 2NH4OH ==>(NH4)2T + 2H2O
mols H2T in 75 g = grams/molar mass = 75/150.087 = about 0.5M
mols H2T used = M x L = 0.5 x 0.025 = 0.0125.
Convert to mols NH4OH using the coefficients in the balanced equation.
mols NH4OH = 2x mols H2T.
Then M NH4OH = mols NH4OH/L NH4OH = ? and that's the same as normality for NH4OH.
You can do it in normality from the start.
H2T = 75g/L.
equivalents = 75g/eq weight = 75/75.043 = about 1M
Then mL x N = mL x N
25 x 1 = 31 x N
N = 25 x 1/31 = ?N
Should be the same answer either way.
I thought about the syntax of the problem and I have surmised that what I thought were two questions was actually one. I'm slow but not dead yet.
75g of tartaric acid in 1 liter is used in a titration. 25mL of tartaric acid solution is titrated against 31mL of NH4OH. What is the normality of the NH4OH?
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