15.00mL of 1.600M NaOH solution exactly neutralizes 20.00mL of an unknown monoprotic acid solution. Calculate the molarity of the acid solution. I got .686M... is this correct?

Answer 1

No, it's not correct.

Liters*molarity = moles

(0.015L)*(1.600moles/L) = 0.024 moles of NaOH

One mole of NaOH would neutralize one mole of the MONOprotic acid, so: (0.024 moles of acid)/(0.020L) = 1.2M

Answer 2

It is found that 24.68 mL of 0.1165 M NaOH is needed to titrate 0.2931 g of an unknown monoprotic acid to the equivalence point Calculate the molar mass of the acid?

Equivalent weight = Molar mass ÷ Total + or - charge

Since Na = +1, Equivalent mass = Molar mass

# of Equivalents = Normality * Liters of solution

# of Equivalents of NaOH = 0.1165 *0.02468 = 0.02878

1 Equivalent of acid will neutralize 1 equivalent of base

So the 0.2931 g of an unknown monoprotic acid = 0.002878 equivalents of acid

Since the acid is monoprotic (1 H = +1 charge),

# of Equivalents = # of moles

So, 0.2931 g of an unknown monoprotic acid = 0.002878 moles

Moles = mass ÷ molar mass

Molar mass = mass ÷ moles

Molar mass = 0.2931 g ÷ 0.002878 = 101.8

Molar mass of unknown acid = 101.8

Equivalent weight = Molar mass ÷ Total + or - charge

Answer 3

Bromthymol blue