How many milliliters of 4.8 M NaOH are needed to exactly neutralize 29 milliliters of 2 M HCl?
1 answer:
Answer:
V₁ = 12.1 mL
Explanation:
Given data:
Volume of NaOH required to neutralize = ?
Molarity of NaOH = 4.8 M
Volume of HCl = 20 mL
Molarity of HCl = ?
Solution:
Chemical equation:
HCl + NaOH → NaCl + H₂O
In neutralization reaction the equal amount of H⁺ from acid and OH⁻ from base react to form water and salt is also formed.
Formula:
M₁V₁ = M₂V₂
M₁ = molarity of NaOH
V₁ = Volume of NaOH
M₂ = molarity of HCl
V₂ = volume of HCl
Now we will put the values.
4.8 M × V₁ = 2 M × 29 mL
4.8 M × V₁ =58 M. mL
V₁ = 58 M. mL / 4.8 M
V₁ = 12.1 mL
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