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Sagot :
Answer:
Explanation:
moles of hydrochloric acid = 50 x 10⁻³ x .75 = 37.5 x 10⁻³ moles in HCl solutions.
moles of NaOH in its solution = 10 x 10⁻³ x 1 moles = 10 x 10⁻³ moles .
10 x 10⁻³ moles of NaOH will neutralise 10 x 10⁻³ moles of HCl .
HCl remaining = (37.5 - 10 ) x 10⁻³ moles = 27.5 x 10⁻³ moles
volume of solution = 50 + 10 = 60 mL = 60 x 10⁻³ L
molarity of remaining HCl = 27.5 x 10⁻³ / 60 x 10⁻³
= .4583 M .HCl
concentration of H⁺ = .4583 M
pH = - log ( .4583 )
= .34 .
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