1. En una solución buffer NH4OH /NH4Cl pH= 9.20 pH + pOH = 14 kb =1,8x10-5
pOH = 14-pH pKb = -log 1,8x10-5
pOH = 14-9.20 = 4.8 pKb = 4,74
Sol. NH4OH = 0,15 M
NH4OH 30% Densidad=0.9 g/mL
[sal]
pOH = pKb + log ---------
[base]
4,8 = 4,74 + log _[sal]____
0,15
4,8 = 4,74 + log [sal] – log [0,15]
Log [sal] = 4,8 -4,74+ (-0.82)
Log [sal] = -0,76
[sal] = 10-0,76
[sal] =0,17
[sal] = 0,17M = NH4Cl V=200 mL
M = n/V
n = M*V = 0,17 molL-1*0,2L = n=3,4x10-2 mol
n =m/PM m = n*PM m = 3,4x10-2 mol * 53,5gmol-1 =1,819 g
[base] = 0,15 M
NH4OH 30% Densidad=0.9 g/mL
M = n/V
n = M*V = 0,15 molL-1*0,2L = n= 3x10-2 mol
n =m/PM m = n*PM m = 3x10-2 mol * 35gmol-1 =1,05 g
100 g ---- 30g NH4OH
x -----1,05 g NH4OH
x = 3,5 g
D = m/V V = m/D V = 3,5g/ 0,9gmL-1 =3,89 mL
