Question

# How many grams of K2HPO4 must be added to 2L of a 0.1 M KH2PO4 SOLUTION...

How many grams of K2HPO4 must be added to 2L of a 0.1 M KH2PO4 SOLUTION TO PRODUCE A PH=7.0 BUFFER? AND CAN YOU IDENTIFY THE ACID, BASE, CONJUGATE BASE AND CONJUGATE ACID AS WELL. PLEASE WORK THE STEP OUT STEP BY STEP

pH = pKa + log[Conjugate base/ Acid]

given values

pH = 7.0 , pKa of acid which is KH2PO4 here = 6.82 (from google)

concentration of Acid = 0.1M

put all these in the above equation

7.0 = 6.82 + log[K2HPO4/0.1]

log[K2HPO4/0.1] = 0.18

[K2HPO4/0.1] = 100.18

[K2HPO4/0.1] = 1.5135

[K2HPO4] = 1.5135 x 0.1

[K2HPO4] = 0.1513 M

now we know the molarity of [K2HPO4] and volume 2L

calculate the moles using molarity M = no of moles / volume in liters

0.1513 = no ofmoles / 2

no of moles = 0.3027 moles

now use the formula moles = mass / molar mass

mass = 0.3027 moles x 174.17 g /moles = 52.7234 grams

Part 2

lets see the equation

KH2PO4 + KOH -----> K2HPO4 + H2O

so here acid is KH2PO4

Base is KOH

conjugate base = K2HPO4

conkugate acid is H2O

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