5. A 22.6 mL sample of 0.105 M calcium hydroxide is titrated to the equivalence

Question

5. A 22.6 mL sample of 0.105 M calcium hydroxide is titrated to the equivalence point with 14.25 mL of hydrobromic acid. Calculate the molarity of the hydrobromic acid solution. 6. How many mL of 0.876 M sulfuric acid would be needed to neutralize 22.4 mL of a 1.25 M solution of aluminum hydroxide? 7. A 23.5 mL sample of sulfurous acid is titrated with 45.0 mL of 0.25 M potassium hydroxide. Calculate the concentration of the sulfurous acid. 8. What volume of 0.45 M phosphoric acid is needed to titrate 17.2 mL of 0.14 M calcium hydroxide?

Explanation / Answer

5)
Balanced chemical equation is:
Ca(OH)2 + 2 HBr ---> CaBr2 + 2 H2O


Here:
M(Ca(OH)2)=0.105 M
V(Ca(OH)2)=22.6 mL
V(HBr)=14.25 mL

According to balanced reaction:
2*number of mol of Ca(OH)2 =1*number of mol of HBr
2*M(Ca(OH)2)*V(Ca(OH)2) =1*M(HBr)*V(HBr)
2*0.105*22.6 = 1*M(HBr)*14.25
M(HBr) = 0.3331 M

Answer: 0.333 M

6)
Balanced chemical equation is:
2 Al(OH)3 + 3 H2SO4 ---> Al2(SO4)3 + 6 H2O


Here:
M(Al(OH)3)=1.25 M
M(H2SO4)=0.876 M
V(Al(OH)3)=22.4 mL

According to balanced reaction:
3*number of mol of Al(OH)3 =2*number of mol of H2SO4
3*M(Al(OH)3)*V(Al(OH)3) =2*M(H2SO4)*V(H2SO4)
3*1.25 M *22.4 mL = 2*0.876M *V(H2SO4)
V(H2SO4) = 47.9 mL

Answer: 47.9 mL

Only 1 question at a time please

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

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