You have a 1.153 g sample of an unknown solid acid, ha, dissolved in enough water to make 20.00 ml of solution. ha reacts with koh(aq) according to the following balanced chemical equation: ha(aq) + koh (aq) --> ka (aq) + h20 (l)if 12.15 ml of 0.715 m koh is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid? what is the molar mass of ha?

The concentration of the unknown acid (HA) is 0.434M

The molar mass of HA is 13.3g/mole

Explanation:

DETERMINATION OF MOLARITY OF THE UNKNOWN ACID

CaVa/CbVb = Na/Nb

From the equation of reaction and at equivalence point, Na = Nb = 1

Therefore, CaVa = CbVb

Va (volume of acid solution) = 20mL = 20/1000 = 0.2L

Cb (concentration of KOH) = 0.715M

Vb (volume of KOH) = 12.15mL

Ca (concentration of acid) = CbVb/Va

Ca = 0.715M × 12.15mL/20mL = 0.434M

DETERMINATION OF MOLAR MASS OF HA

Number of moles of acid = concentration of acid × volume of acid solution in liters = 0.434 × 0.2 = 0.0868mole

Molar mass of HA = mass/number of moles = 1.153g/0.0868mole = 13.3g/mole

i dont know the answe

it is stable as it is in group 0 which have full outer shells of electrons