What is the molarity of a potassium hydroxide solution if 38.65 mL of the KOH solution is required to titrate 25.84 mL of 0.1982 M hydrochloric acid solution?
0.000197
To determine the molarity of the potassium hydroxide (KOH) solution, we can use the equation:
M1V1 = M2V2
Where:
M1 = molarity of the hydrochloric acid (HCl) solution
V1 = volume of the HCl solution
M2 = molarity of the KOH solution
V2 = volume of the KOH solution
In this case, we know:
M1 = 0.1982 M (given)
V1 = 25.84 mL (given)
V2 = 38.65 mL (given)
Let's substitute these values into the equation and solve for M2:
M1V1 = M2V2
(0.1982 M)(25.84 mL) = M2(38.65 mL)
Now, let's solve the equation:
M2 = (0.1982 M)(25.84 mL) / 38.65 mL
M2 ≈ 0.1328 M
Therefore, the molarity of the potassium hydroxide solution is approximately 0.1328 M.
To find the molarity (M) of the potassium hydroxide (KOH) solution, we can use the formula:
M1 × V1 = M2 × V2
Where:
M1 is the molarity of the hydrochloric acid (HCl) solution
V1 is the volume of the HCl solution used for titration
M2 is the molarity of the KOH solution
V2 is the volume of the KOH solution used for titration
In this case:
M1 = 0.1982 M (given)
V1 = 25.84 mL (given)
V2 = 38.65 mL (given)
Now, we can plug in these values into the formula:
(0.1982 M) × (25.84 mL) = (M2) × (38.65 mL)
To solve for M2, we rearrange the equation:
M2 = (0.1982 M × 25.84 mL) / 38.65 mL
Let's calculate it step by step using these values:
(M2) = (0.1982 M × 25.84 mL) / 38.65 mL
(M2) = 5.1225288 / 38.65
(M2) ≈ 0.1326 M
Therefore, the molarity of the potassium hydroxide (KOH) solution is approximately 0.1326 M.
molesbase=molesacid
M*volumebase=volumeacid*concentrationHCl
molarity= (volumeacid/volume base)*.1982M