A sample of 0.64 g of an unknown diprotic acid requires 15.6 mL of 1.00 M KOH so

Question

A sample of 0.64 g of an unknown diprotic acid requires 15.6 mL of 1.00 M KOH solution to completely neutralize the acid. What is the molar mass (in g/mol) of this diprotic acid

If 42.0 mL of 0.100 M NaOH solution is required to completely react with 10.0 mL of a phosphoric acid solution, what is the concentration of the phosphoric acid?

Which of the following acidic solutions require the greatest volume of 0.10 M NaOH to complete the neutralization reaction?

A

20.0 mL 0.20 M CH3COOH

B

30.0 mL 0.10 M H3PO4

C

40.0 mL 0.10 M H2SO4

D

40.0 mL 0.15 M HNO2

Explanation / Answer

1)

mol of KOH reacting = M*V

= 1.00 M * 0.0156 L

= 1.56*10^-2 mol

Since the acid is diprotic, 1 mol of acid will give 2 mol of H+

So,

mol of acid reacting = (1/2)*moles of KOH

mol of acid reacting = (1/2)*1.56*10^-2 mol

mol of acid reacting = 7.80*10^-3 mol

Now use:

mol of acid = mass / molar mass

7.80*10^-3 mol = 0.64 g / molar mass

molar mass = 82.1 g/mol

Answer: 82.1 g/mol

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