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Answer:

Volume of NaOH  is 125 mL

Explanation:

Given;

volume of HCL, Va = 250 mL

concentration of HCL, Ca = 5 M

volume of  NaOH, Vb = ?

concentration NaOH, Cb = 10 M

HCl + NaOH -----------> NaCl + H₂O (neutralization reaction)

na (number of moles of Acid) = 1

nb (number of moles of base) = 1

na = nb ⇒ CaVa = CbVb

where

Ca is acid concentration

Cb is base concentration

Va is volume of acid

Vb is volume of base

CaVa = CbVb

Vb = (CaVa) / (Cb)

Vb = (5 x 250) / (10)

Vb = 125 mL

Therefore, the volume of NaOH that will completely neutralize HCl is 125 mL

Answer:

Volume of NaOH is 0.125 L

Explanation:

HCl + NaOH --------> NaCl + H2O

Using the molarity formulae below:

MA VA / MB VB = nA / nB

MA = Molar volume of acid = 5.0 M

VA = volume of acid = 250 mL = 250 / 1000 L = 0.25 L

MB = molar volume of base = 10 M

nA = number of mole of acid = 1

nB = number of mole of base = 1

VB = ?

Vb = MA VA nB/ nA MB

VB = 5 * 0.25 * 1 / 1 * 10

VB = 1.25 / 10

VB = 0.125 L

The volume of 10 M NaOH at 250 mL to be used to neutralize 5M of HCl is 0.125 L

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