A buffer solution is composed of 1.067 g of KH2PO4 and 4.049 g of NaHPO4. ( for
ID: 501150 • Letter: A
Question
A buffer solution is composed of 1.067 g of KH2PO4 and 4.049 g of NaHPO4. ( for dihydrogen phosphate ion is 6.2x10^-8.)
What is the pH of the buffer solution? pH =
What mass of KH2PO4 must be added to decrease the buffer solution pH by 0.10 unit from the value calculated in part a?
Mass = g
Please explain and write legibly! Thanks! :)
You dissolve 0.106 rofNaOR in 200 Lefabuner solution that has II,PO. HPO" 0151 M What de pHerde solution before adding NaOll ion is 6.2 x 10 pH- What is pHof the soluton after allang NaOH Ander version 5 attempts remainingExplanation / Answer
a) Let be volume of buffer be 1litre
Molar mass of KH2PO4 = 136.09
Mass = 1.067g
No of mole = 1.067/136.09=
[KH2PO4]= 0.00784mole/litre = 0.00784M
Molar mass of NaHPO4 = 118.99g
Mass of NaHPO4 = 4.049g
No of mole NaHPO4= 4.049/118.99= 0.03403
[NaHPO4] = 0.003403mol/litre = 0.003403M
Ka = 6.2 ×10^-8
pKa =-log(Ka)= 7.21
Henderson equation is
pH = pKa + log ( [A-]/[HA])
pH = 7.21+ log (0.03403/0.00784)
= 7.21 + 0.64
= 7.85
b) After addition of KH2PO4 ,pH = 7.85 -0.10 = 7.75
Applying Henderson equation
7.75 = 7.21 + log ( 0.03403/[KH2PO4])
log [KH2PO4] =-2.008
[KH2PO4] = 0.00982M = 0.00982 mole /litre
Difference mole = 0.00982-0.00784= 0.00198mole
Molar mass of KH2PO4 = 136.09g
Therefore , Mass of KH2PO4 = 136.09× 0.00198= 0.269g
Therefore 0.269g of KH2PO4 must be added
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.