Respuesta :
Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]
Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess.
Follow the steps as
1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.
n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-
Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess
n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.
Total volume= V acid + V base= 28 ml + 60 ml = 98 ml
Molarity of OH- ions= mole / Vtotal = 0.92/98= 0.009 M
The answer is 0.009 M.
Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess.
Follow the steps as
1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.
n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-
Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess
n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.
Total volume= V acid + V base= 28 ml + 60 ml = 98 ml
Molarity of OH- ions= mole / Vtotal = 0.92/98= 0.009 M
The answer is 0.009 M.
Answer: 0.52 M
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of [tex]HCl[/tex] solution = 0.10 M
Volume of [tex]HCl[/tex] solution = 28 mL = 0.028 L
Putting values in equation 1, we get:
[tex]\text{Moles of} HCl={0.10}\times{0.028}=0.0028moles[/tex]
Molarity of [tex]Sr(OH)_2[/tex] solution = 0.10 M
Volume of [tex]Sr(OH)_2[/tex] solution = 60 mL = 0.060 L
Putting values in equation 1, we get:
[tex]\text{Moles of} Sr(OH)_2={0.10}\times{0.060}=0.006moles[/tex]
[tex]2HCl+Sr(OH)_2\rightarrow SrCl_2+2H_2O[/tex]
According to stoichiometry :
2 moles of [tex]HCl[/tex] neutralize 1 mole of [tex]Sr(OH)_2[/tex]
Thus 0.0028 moles of [tex]HCl[/tex] neutralize=[tex]\frac{1}{2}\times 0.0028=0.0014moles[/tex] of [tex]Sr(OH)_2[/tex]
Thus (0.006-0.0014) moles = 0.046 moles of [tex]Sr(OH)_2[/tex] are left in 88 ml of solution.
Concentration of [tex][OH^-][/tex] will be =[tex]\frac{moles}{\text {Volume in L}}=\frac{0.046}{0.088}=0.52M[/tex]
Thus the concentration of [tex][OH^-][/tex] in the resulting solution is 0.52 M
