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9 months ago

what volume of a 0.138 m potassium hydroxide solution is required to neutralize 26.0 ml of a 0.205 m nitric acid solution?

1 answer:

4 0

A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is ( $M1V1 = M2V2$ ).

Utilize the titration method of $M1V1 = M2V2$ in view that we're given the concentrations of every compound and the quantity of $KOH$ . Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.

M1 = initial mass
V1= initial volume
M2 = final mass
V2= final volume

$(M1V1 = M2V2)$
(0.138)(V1) = (0.205)x(26.0)
V2=(0.205)x(26.0)\ 0.138
V2 = 47.10 M/L
The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L

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2 years ago

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2 years ago

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andreyandreev [35.5K]

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<u>Explanation:</u>

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$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the NaOH.

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where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

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We are given:

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