25.0 mL of 0.100 M NaOH requires 17.2 mL of a sulfuric acid solution to neutralize. What is the M of the sulfuric acid solution? (report answers in M)
2007-02-11 00:38:39 · 4 answers · asked by ? 2 in Science & Mathematics ➔ Chemistry
H2SO4+2NaOH--> Na2SO4 +H2O
You see that the ratio NaOH /H2SO4 = 2
So to neutralize 25mL of 0.1M NaOH , you need 12.5mL of H2SO4
if x is the molarity of 17.2mL of H2SO4, you can write
17.5x = 12.5*0.1 or
x = 12.5*0.1/17.5 =0.071
The solution is 0.071M
2007-02-11 00:52:33 · answer #1 · answered by maussy 7 · 1⤊ 0⤋
direct formula
M1V1/n1=M2V2/n2
let us take M1= Molarity of NaOH
V1= Volume of NaOH
n1= Number of moles of NaOH[ACCORDING TO THE BALANCED EQUATION ]
M2= Molarity of H2SO4
V2= Volume of H2SO4
n2= Number of moles of H2SO4 [ACCORDING TO THE BALANCED EQUATION ]
The balanced equation is
2NaOH+H2SO4--------->Na2SO4+2H2O
n1=2
n2=1
M1=0.1
M2=?
V1=25ml
V2= 17.2ml
Now substitute the values in the formula
0.1 x 25/2 = M2 x 17/ 1
M2= 0.1X 12.5/17
M2= 0.0735
0.0735M IS THE MOLARITY OF H2SO4
2007-02-11 02:32:57 · answer #2 · answered by manidhar 2 · 0⤊ 0⤋
2NaOH + H2SO4 ---> Na2SO4 + 2H2O
12.5cm^3 are needed... (2:1) ratio
M = 12.5*0.1/17.5= 0.0714M
2007-02-11 00:59:07 · answer #3 · answered by SS4 7 · 0⤊ 0⤋
M1V1 = M2V2
17.2 XM1 = 25X 0.1
M1 = 25X0.1/17.2
= 2.5/17.2
= 0.145
M = 0.145/2 = 0.072 M
2015-08-09 21:24:35 · answer #4 · answered by Rameshwar 7 · 0⤊ 0⤋