Respuesta :
Answer:
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
The molar mass of unknown HA solution has been 132.8 g/mol.
Titration has been given as the neutralization reaction for acid and base to result in the formation of salt and water.
The balanced chemical equation for the reaction has been:
[tex]\rm HA\;+\;KOH\;\rightarrow\;KA+\;H_2O[/tex]
Computation for the molar mass of HA
The molarity of the unknown acid solution has been given as:
[tex]M_1V_1=M_2V_2[/tex]
Where, the molarity of KOH solution, [tex]M_2=0.715\;\rm M[/tex]
The volume of the KOH solution, [tex]V_2=12.15\;\rm mL[/tex]
The volume of the HA solution, [tex]V_1=20.0\;\rm mL[/tex]
Substituting the values for the molarity of HA, [tex]M_1[/tex]
[tex]M_1\;\times\;20=0.715\;\times\;12.15\\M_1=\dfrac{0.715\;\times\;12.15}{20} \\M_1=0.434\;\rm M[/tex]
The concentration of the HA solution has been 0.434 M.
The molar mass from molarity has been given by:
[tex]\rm Molar \;mass=\dfrac{Mass}{Molarity}\;\times\;\dfrac{1000}{Volume\;mL}[/tex]
Substituting the values for the molar mass of HA:
[tex]\rm Molar \;mass=\dfrac{1.153}{0.434}\;\times\;\dfrac{1000}{\;20}\\Molar\;mass=2.656\;\times\;50\\Molar \;mass=132.8\;g/mol[/tex]
The molar mass of unknown HA solution has been 132.8 g/mol.
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