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3HCl(aq) + Al(OH)3(aq) = AIC13(aq) + 3H2O(1)
Based on the reaction shown above, what volume of 0.246 M
HCl is needed to titrate a 200 mL of 0.048 M Al(OH)3 solution?
mL

Respuesta :

Answer:

117.07 ml

Explanation:

Right on ck-12

Eduard22sly

The volume of 0.246 M HCl needed to titrate  200 mL of 0.048 M Al(OH)₃ solution is 117 mL

Balanced equation

3HCl + Al(OH)₃ —> AlCl₃ + 3H₂O

From the balanced equation above,

  • The mole ratio of the acid, HCl (nA) = 3
  • The mole ratio of the base, Al(OH)₃ (nB) = 1

How to determine the volume of HCl

  • Molarity of acid, HCl (Ma) = 0.246 M
  • Molarity of base, Al(OH)₃ (Mb) = 0.048 M
  • Volume of base, Al(OH)₃ (Vb) = 200 mL
  • Volume of acid, HCl (Va) =?

MaVa / MbVb = nA / nB

(0.246 × Va) / (0.048 × 200) = 3

(0.246 × Va) / 9.6 = 3

Cross multiply

0.246 × Va = 9.6 × 3

Divide both side by 0.246

Va = (9.6 × 3) / 0.246

Va = 117 mL

Learn more about titration:

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