Question

28 ml of 0.10 m hcl is added to 60 ml of 0.10 m sr(oh)2. determine the concentration of oh− in the resulting solution.

Answer 1

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.

Answer 2

Answer: 0.52 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$     .....(1)

Molarity of $HCl$ solution = 0.10 M

Volume of $HCl$ solution = 28 mL = 0.028 L

Putting values in equation 1, we get:

$\text{Moles of} HCl={0.10}\times{0.028}=0.0028moles$

Molarity of $Sr(OH)_2$ solution = 0.10 M

Volume of $Sr(OH)_2$ solution = 60 mL = 0.060 L

Putting values in equation 1, we get:

$\text{Moles of} Sr(OH)_2={0.10}\times{0.060}=0.006moles$

$2HCl+Sr(OH)_2\rightarrow SrCl_2+2H_2O$  

According to stoichiometry :

2 moles of $HCl$ neutralize 1 mole of $Sr(OH)_2$

Thus 0.0028 moles of $HCl$ neutralize=$\frac{1}{2}\times 0.0028=0.0014moles$ of $Sr(OH)_2$

Thus (0.006-0.0014) moles = 0.046 moles of $Sr(OH)_2$  are left in 88 ml of solution.

Concentration of $[OH^-]$ will be =$\frac{moles}{\text {Volume in L}}=\frac{0.046}{0.088}=0.52M$

Thus the concentration of $[OH^-]$ in the resulting solution is 0.52 M