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Sagot :
Answer:
Explanation:
Mol weight of HCl = 36.5
.90 g of HCl = .90 / 36.5 mole = .02465 mole
HCl = H⁺ + Cl⁻
1 mole 1 mole
1 mole of HCl gives 1 mole of H⁺
.02465 mole of HCl will give .02465 mole of H⁺
.02465 mole of H⁺ in 5 litre soln
concentration of H⁺ = .02465 / 5 = 4.93 x 10⁻³ M
pH = - log ( 4.93 x 10⁻³ )
= 3 - log 4.93
= 3 - .693
= 2.307
= 2.31
NaOH :
Mol weight of NaOH = 40
.50 g of HCl = .50 / 40 mole = .0125 mole
NaOH = Na⁺ + OH⁻
1 mole 1 mole
1 mole of NaOH gives 1 mole of OH⁻
.0125 mole of NaOH will give .0125 mole of OH⁻
.0125 mole of OH⁻ in 4 litre soln
concentration of OH⁻ = .0125 / 4 = 3.125 x 10⁻³ M
pOH = - log ( 3.125 x 10⁻³ )
= 3 - log 3.125
= 3 - .495
= 2.505
pH = 14 - pOH
= 14 - 2.505
= 11.495 .
= 11.50
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