Answer on Question #48803, Chemistry, Other
Task:
Calculate the volume of a 0.500 M NaOH solution needed to neutralize (titrate to
the endpoint):
a)10.0 mL 0.300M HCl
b)10.0 mL 0.200M H2SO4
c) 4.08 g KHP (molar mass 204 g/mol)
Answer:
NaOH+HCl=NaCl+H2O
CM ( NaOH ) V ( NaOH )  CM ( HCl ) V ( HCl )
V ( NaOH ) 
0,300 10, 0
 6, 00 ml
0,500
2NaOH+ H2SO4=Na2SO4+2H2O
2 ( NaOH )   ( Na2 SO4 )
 ( NaOH ) 
 ( Na2 SO4 )
2
CM ( NaOH )  V ( NaOH ) 
V ( NaOH ) 
CM ( Na2 SO4 )  V ( Na2 SO4 )
2
0, 200 10, 0
 2 ml
2  0,500
NaOH+KHP= KNaP + H2O
m( NaOH )
M (( NaOH ))
 ( NaOH )   ( KNaP)
4, 08
 ( NaOH ) 
 0, 02 mol
204
СM ( NaOH ) V ( NaOH )   ( NaOH )
 ( NaOH ) 
V ( NaOH ) 
 ( NaOH )
СM ( NaOH )

0, 02
 0, 04 ml
0,500
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