Question

A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resulting solution?A) 0.10 MB) 0.19 MC) 0.28 MD) 0.40 ME) 0.55 M

Answer 1

The molar concentration of OH- is 0.28M (Option C)

Step-by-step explanation:

Step 1: Data given

Volume of 0.25 M KOH = 40.0 mL = 0.04 L

Volume of 0.15 M Ba(OH)2 = 60.0 mL = 0.06 L

Step 2: Calculate moles of KOH

Moles KOH = Molarity KOH * volume

Moles KOH = 0.25 M * 0.04 L = 0.01 moles

Step 3: Calculate moles of Ba(OH)2

Moles Ba(OH)2 = 0.15 M * 0.06 L

Moles Ba(OH)2 = 0.009 moles

Since We have twice as much OH in Ba(OH)2 as we have in KOH

The quantity of moles in Ba(OH)2 is 0.009 * 2 = 0.018

Step 4: Calculate the total number of moles

The total amount of OH is 0.018 mol + 0.010 mol = 0.028 mol

Step 5: Calculate molarity

Molarity = Number of moles / volume

⇒ with nummber of moles = 0.028 moles

⇒ with volume = 40 + 60 = 100 mL = 0.100 L

The total volume is 100ml = 0.1L

Molarity = 0.028 mol/ 0.1 L = 0.28 M

The molar concentration of OH- is 0.28M

Answer 2

C) 0.28 M

Step-by-step explanation:

Considering:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Moles =Molarity * {Volume\ of\ the\ solution}`

Potassium hydroxide will furnish hydroxide ions as:

`KOH\rightarrow K^(+)+OH^-`

Given :

For Potassium hydroxide :

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

`Moles =0.25 * {40.0* 10^(-3)}\ moles`

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

`Ba(OH)_2\rightarrow Ba^(2+)+2OH^-`

Given :

For Barium hydroxide :

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

`Moles =2* 0.15 * {60.0* 10^(-3)}\ moles`

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Molarity_(OH^-)=(0.028 )/(100* 10^(-3))`

The final concentration of hydroxide ion = 0.28 M