Question

x5.a。. 21 2ic Sample of an organic ampound was burned in a stream of oygenyand the coz pradcnd was clleetal in a Solutsion (hariwn 2A o.4000 Samp 4l.25onL of barium akloride Selution Gladale the mobritg Qadde the PH-e the Solut, lon after tee adlition申O,25mL of the acid. (ta HC Ns Gaxilo ) 2o muof o.2oM HCiOy to 50.0oml Solutron in uce | hydrazine is 0.08m (kb:9,salo). th her&e aldition哻loat e Se luti tion otlo or in empatible ca mbinetion was titnetl with o.1202 M Hal. The Yolume faced nealad to reach tee wso 15.67 mc. The Volume neeid totitrate 2s mt Sample to Hhe bro mo aresel reen eud point was 42 13 ml Deluce 4Rc cempos ition ofthe mitture anl Calalnte the Concestro f ench base phenophthalein end peint Scanned with CamScan

Answer 1

1)

Organic compound -------> BaCO3

1 mole of BaCO3 requires 1 mole of C

197.34 g of BaCO3 requires 12 g of C

0.6006 g of BaCO3 requires 12/197.34*0.6006 g of C

                         = 0.0365 g of C

percentage of C = 0.0365/0.2121*100 = 17.21%

2)

Na2SO4 + BaCl2 -----> BaSO4 + 2NaCl

number of moles of Na2SO4 = 0.4/142.04 = 0.00282 moles

but given is 96.4% of Na2SO4 is present in the sample so,

moles of the given sample = 0.00282*100/96.4 = 0.002925 moles

1 mole of Na2SO4 require 1 mole of BaCl2 solution,

molarity of the solution = 0.002925/41.25*1000 = 0.071 mol/L

3)

number of moles of NaCN = 50 * 0.05 = 2.5 mmol

befor any addition of HCl,

pH = 1/2(pKa + logC)

pH = 1/2(-log(6.2*10^-10) + log(2.5*10^-3))

pH = 3.3028

-logH+ - 3.3028

[H+] = 10^-3.3028

[H+] = 4.98*10^-4

number of moles of HCl = 25*0.1 = 2.5 mmol = 2.5 *10^-3 mol

total number of moles of [H+] = (2.5*10^-3)+(4.98*10^-4) = 0.003 moles

total concentration of H+ ions = 0.003*1000/(50+25) = 0.04

pH = -log(H+)

pH = -log(0.04)

pH = 1.398