mada4pethcha
Answered

In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?

Answer :

Hagrid
M(HCl) * V(HCl) = M(NaOH) * V(NaOH)

M(HCl) = 0.35
V(HCl) = 45mL
M(NaOH)= 0.35

now, solne for V(NaOH) by putting these values in the above equation.
M(HCl) * V(HCl) = M(NaOH) * V(NaOH)

0.35 * 45 = 0.35 * V(NaOH)

V(NaOH) = 45 mL

how it cumz is

acc to the law of volumetric anlysis

V1N1 = V2N2

take V1 as vol of HCl = 0.35 M

V2 as vol of NaOH = 0.35M

N1 of HCl = 45.0

N2 of NaOH = ?

N2 = V1 x N1

V2

= 0.35 x 45.0

----------------

0.35

= 45.0

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