How many grams of NaOH are required to neutralize 15.0m of .85 M HCl

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How Many mL of 0.15 M NaOH solution are required to neutralize 35.00 mL of 0.22 M HCl?

Titration of a 25.00 mL of KOH required 200.00 mL of 0.0050 M acetic acid, CH3COOH. What is the molarity of the KOH solution?

Obtain the moles of HCl first.
0.85 x 15 / 1000 = 0.01275mols of HCl
Mm of NaOH
Na = 22.989768
O = 15.9994
H = 1.00794
NaOH = 22.989768 + 15.9994 + 1.00794 = 39.997108g/mol
Grams = Moles x Mm of NaOH
0.01275 x 39.997108 = 0.51g of NaOH

2)
Molarity x volume = moles
0.22 x 35.00 / 1000 = 0.0077mols of HCl
mL = moles / molarity
0.0077 / 0.15 = 0.0513L or 51mL

3)
0.0050 x 200 / 1000 = 0.003mols of CH3COOH
0.003 / 25.00 / 1000 = 0.12M of KOH

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