Question

What is the concentration of an hbr solution if 12.0 ml of the solution is neutralized by 15.0 ml of a 0.25 m koh solution?

Answer 1

The concentration of the HBr solution is approximately 0.3125 M.

Step-by-step explanation:

To find the concentration of the HBr solution, we can use the equation:

M1V1 = M2V2

where M1 is the concentration of the KOH solution (given as 0.25 M), V1 is the volume of the KOH solution (given as 15.0 mL), M2 is the concentration of the HBr solution (what we're trying to find), and V2 is the volume of the HBr solution (given as 12.0 mL).

Plugging in the values:

(0.25 M)(15.0 mL) = M2(12.0 mL)

Simplifying:

3.75 = 12M2

Dividing both sides by 12:

M2 = 3.75/12 ≈ 0.3125 M

Therefore, the concentration of the HBr solution is approximately 0.3125 M.

Answer 2

the balanced equation for the reaction between KOH and HBr is as follows;
KOH + HBr --> KBr + H₂O
stoichiometry of KOH to HBr is 1:1
number of KOH moles reacted - 0.25 mol/L x 0.015 L = 0.00375 mol
according to molar ration
number of KOH moles reacted = number of HBr moles reacted
number of HBr moles reacted - 0.00375 mol
if 12 mL of HBr contains - 0.00375 mol
then 1000 mL of HBr contains - 0.00375 mol / 12 mL x 1000 mL = 0.313 mol
therefore molarity of HBr is 0.313 M