A sample of a crude (impure) grade of barium hydroxide is sent to the lab to be

Question

A sample of a crude (impure) grade of barium hydroxide is sent to the lab to be tested for its barium content. A sample of this crude solid is dissolved and diluted to 700 mL with water. A 90 mL sample of the solution is titrated with a 0.9925 M of nitric acid solution and requires 58.43 mL to be neutralized. a. How many mols of nitric acid were needed to neutralize the barium hydroxide? b. What was the molarity of barium hydroxide before the titration? c. How many mols of barium hydroxide were in the original sample? d. What is the percent barium hydroxide in the crude material if the total sample weighed 40.002 gram.

Explanation / Answer

Ans 1.

a)

Number of moles of HNO3 = molarity x volume in L

=0.9925 x 0.05843

= 0.058 moles

b)The balanced reaction is

Ba(OH)2 + 2HNO3 = Ba(NO3)2 + 2H2O

So by using th formula

M1 V1 / n1 = M2 V2 / n2

where M is molarity , V is volume and n is number of moles

putting the values in the formula

(M1 x 90) / 1 = (0.9925 x 58.43) / 2

M1 = 0.3222 M

c) Number of moles of Ba(OH)2 = molarity x volume in Litres

= 0.3222 x 0.700

= 0.2255 moles

d) Molar mass of Ba(OH)2 = 171.34 g/mol

So 0.2255 moles will weigh 0.2255 x 171.34 = 38.64 grams

Percent barium hydroxide = (actual amount / total amount ) x 100

= (38.64 / 40.002) x 100

= 96.6 %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---