h@yooog sa ple that ass ged q64% Mzsoy (142\'oeleld)re I irde ?·25mL of barium c

Question

h@yooog sa ple that ass ged q64% Mzsoy (142'oeleld)re I irde ?·25mL of barium chloride Sdution . aladtetke Moan ty of A 50.00 ml aliquotaf o.o5M ?? ) is titrated with o.i0HHC, Clastate the pH the Sout ion after the addition Po, 25 m aleulale the PH of the Selution after he addition of lo a 2-5 mL4o, 20 M HC?Oy to 50.00 mL Solution in wecch nd hydrazine is o.os m (kh 9s2xi) A a- ml Schetion Contouning Nao?y Nazco aned NaHCo,alone or in cempatible Co mbinetion was -t itnetl with o.1202 MHC The Yalume faed neeed to reach the phenophthealein el pei woo 15-67 m. The Volume needet totitrate 2s m Sample to tee bromoresol reen e point was 42-13 ml Deduce the Composition oftte Mitture an alaite tte Concet rat f ench base

Explanation / Answer

1) Mass of BaCl2 = volume x density

Volume of BaCl2 = 41.25 mL

Density of  BaCl2 =  3.86 g/cm³

  Mass of BaCl2 = 41.25 mL x 3.86 g/cm³

Mass of BaCl2 = 159,225 g

Molar mass  BaCl2 = 208.23 g/mol

Moles of BaCl2 = mass/ molar mass

=  159,225 g / 208.23 g/mol

Moles of BaCl2 = 0.7647 mol

Molarity of BaCl2 =  moles of BaCl2 / L solution

=  0.7647 / 0.04125

Molarity of BaCl2 = 18.5382 M

2)

pKa = -log (Ka)

pKa = -log (6.2 x 10-10)

pKa = 9.2

M1V1=M2V2

0.05M x 50.00 mL = M2 x 0.25 mL

M2 = 10 M

pH = pKa + log ( [A-] / [HA] )

[H-] = 0.10-10 = 9.9 M

[HA] = 0.10 +10 = 10.1 M

pH = 9.2 + log ( [9.9 M] / [10.1 M] )

= 9.2 + (-0.008686)

pH = 9.19

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.