A40.0 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

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

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

it is a good conductor

explanation:

the answer to your question is c