If K3PO4= 0.250M, how many grams of K3PO4 are in 750.0ml of solution? Remember that M is the same as mol/L. Answer: 39.8g
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<u>Answer:</u> The pH of the solution after addition of KOH is 3.84
<u>Explanation:</u>
We are given:
pH of buffer = 3.75
of formic acid = 3.75
Using Henderson-Hasselbalch equation for formate buffer:
Putting values in above equation, we get:
We are given:
Concentration of formate buffer = 0.100 M
As, the volume of buffer is the same. So, the concentration is taken as number of moles of formate ions as well as formic acid
To calculate the number of moles for given molarity, we use the equation:
Molarity of KOH = 1.00 M
Volume of solution = 5 mL
Putting values in above equation, we get:
The chemical reaction for formic acid and KOH follows the equation:
Initial: 0.05 0.005 0.05
Final: 0.045 - 0.055
Volume of solution = 500 + 5 = 505 mL = 0.505 L (Conversion factor: 1 L = 1000 mL)
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
We are given:
= negative logarithm of acid dissociation constant of formic acid = 3.75
pH = ?
Putting values in above equation, we get:
Hence, the pH of the solution after addition of KOH is 3.84