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A chemist titrates 0.200 M NaOH, strong base, with 50.00 ML of 0.150 M HCI, strong acid. How many mL of NaOH will be required to titrate to the endpoint

Respuesta :

[tex](normality \: of \: acid)×(volum \: of \: acid) = (normality \: of \: base)×(volum \: of \: base)[/tex]

0.15N × 50mL = 0.2N × (Vbase)

75mL = Volum of base

samueladesida43

37.5mL of NaOH will be required to titrate 0.200 M NaOH, strong base, with 50.00 ML of 0.150 M HCI, strong acid to the endpoint.

How to calculate volume?

The volume of a solution can be calculated using the following formula:

C1V1 = C2V2

Where;

  • C1 = initial concentration
  • C2 = final concentration
  • V1 = initial volume
  • V2 = final volume

  • C1 = 0.200M
  • C2 = 0.150M
  • V1 = ?
  • V2 = 50mL

0.2 × V1 = 0.150 × 50

0.2V1 = 7.5

V1 = 7.5/0.2

V1 = 37.5mL

Therefore, 37.5mL of NaOH will be required to titrate 0.200 M NaOH, strong base, with 50.00 mL of 0.150 M HCI, strong acid to the endpoint.

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