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stealth61 [152]
3 years ago
6

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?

Chemistry
2 answers:
Rom4ik [11]3 years ago
8 0
<span>M(HCl) * </span><span>V(HCl) </span>= <span>M(NaOH) * </span><span>V(<span>NaO<span>H)
</span></span></span>
M(HCl) = 0.35
<span>V(HCl) = 45mL
</span>M(NaOH)= 0.35

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

<span>0.35 * 45 = 0.35 * V(NaOH)</span>

<span>V(NaOH) = 45 mL</span>

stich3 [128]3 years ago
6 0

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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