A 1.0L buffer initially contains 0.25mol of NH3 and 0.25mol of NH4Cl. In order to adjust the buffer pH to 8.75 [Solved]

0.81 3.24x = 0.25 x
0.56 = 4.24 x

[NH3]= 0.25mol/ 1.0 L =0.25 M
[MH4 ]= 0.25 mol/ 1.0 L = 0.25 M

mass HCl needed = 0.132 mol x 36.461 g/mol=4.82 g

pOH = 4.74 log 0.25/0.25 = 4.74
pOH = 14 4.74 =9.26
To adjust the buffer pH to 8.75 we must add HCl

3.24 =0.25 x/0.25-x

mass HCl needed = 0.132 mol x 36.461 g/mol=4.82 g

5.25 4.74=0.51

3.24 =0.25 x/0.25-x

5.25 = 4.74 log [NH4 ]/ [NH3]

0.81 3.24x = 0.25 x
0.56 = 4.24 x

x = 0.132 M
moles HCl needed = 0.132 M x 1.0 L= 0.132

0.81 3.24x = 0.25 x
0.56 = 4.24 x

This post is last updated on hrtanswers.com at Date : 1st of September – 2022