rlwatkins2018 rlwatkins2018
  • 01-02-2020
  • Chemistry
contestada

What volume (mL) of 1.50 × 10-2 M hydrochloric acid can be neutralized with 115 mL of 0.244 M sodium hydroxide?

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Eduard22sly
Eduard22sly Eduard22sly
  • 02-02-2020

Answer: 1870. 67mL

Explanation:

NaOH + HCl —> NaCl + H2O

nA = 1, Ca = 0.015M, Va =?

nB = 1 Cb = 0.244 M, Vb = 115 mL

CaVa / CbVb = nA/nB

(0.015xVa)/(0.244x115) = 1/1

Va = (0.244x115) /0.015 = 1870. 67mL

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gbustamantegarcia054 gbustamantegarcia054
  • 02-02-2020

Answer:

Vacid = 1870.66 mL

Explanation:

neutralization point:

  • (V*C)acid = (V*C)base

⇒ Vacid = ((0.115 L)*(0.244 M))/(1.50 E-2 M)

⇒ Vacid =  1.8706 L

⇒ Vacid = 1870.66 mL

Answer Link

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