Respuesta :
Answer:
Explanation:
Sr(OH)₂.+ 2HCl = SrCl₂ + 2H₂O
Moles of HCl in 28mL of .10 M HCl = .028 x .1 = .0028 moles .
Moles of Sr(OH)₂ in 60mL of .10 M Sr(OH)₂ = .060 x .1 = .0060 moles
2 moles of HCl reacts with 1 mole of Sr(OH)₂
.0028 moles of HCl reacts with .0014 mole of Sr(OH)₂
moles of Sr(OH)₂ remaining = .0060 - .0014 = .0046 moles .
Sr(OH)₂ = Sr⁺ + 2OH⁻
1 mole 2 mole
.0046 .0092
Total volume of solution = 88 mL .
88 mL of solution contains .0092 moles of OH⁻
concentration of OH⁻ = .0092 / .088
= .1045 M .
When Strontium hydroxide, Sr(OH)₂ will react with hydrochloric acid (HCl) then the reaction formed will be:
Sr(OH)₂+ 2 HCl ⇒ SrCl₂ + 2 H₂O
The concentration of OH in the resulting solution will be 0.1045 M.
This can be calculated by:
- Molarity (HCl) = 0.10 M
- Volume (HCl) = 28 mL
- Molarity (Sr(OH)₂) = 0.10 M
- Volume (Sr(OH)₂) = 60.0 mL
[tex]\text{Molarity} = \dfrac{\text{moles}}{\text{volume of the solution in litres}}[/tex]
Moles of HCl = 0.028 x 0.1 = 0.0028 moles
Moles of Sr(OH)₂ = 0.060 x 0.1 = 0.0060 moles
- 2 moles of hydrochloric acid will react with 1 mole of Sr(OH)₂
- So, 0.0028 moles of HCl will react with 0.0014 mole of Sr(OH)₂
Remaining moles of Sr(OH)₂ = 0.0060 - 0.0014 = 0.0046 moles
Sr(OH)₂ ⇒ Sr⁺ + 2OH⁻
1 mole 2 mole
0.0046 0.0092
Total volume of solution = 88 mL
88 mL of solution contains 0.0092 moles of OH⁻
Hence, concentration of OH⁻ :
[tex]= \dfrac{0.0092 }{0.088}[/tex]
= 0.1045 M
Therefore, 0.1045 M will be the concentration of OH in the solution.
To learn more about molarity and concentration follow the link:
https://brainly.com/question/15532279
