I will do the third one.
NH4OH: .010l*.3=.0030 moles
HCl: .030*.1=.003 moles
they are all react, no excess acid or base. So, pH=7
the fourth one:
moles base: .0030
moles acid: .040*.1=.004, or an excess of acid of .001moles
volume+ 10ml+40ml=.050 liters
concen acid: .001/.050=.02M
pH= -log(.02)= 1.70
10.0 mL of a 0.300M NH3 solution is titrated with a 0.100M HCl solution. Calculate the pH after the following additions of HCl.
0.00mL
10.00mL
30.00mL
40.00mL
1 answer
1 answer