Procedure for titrating an acid against a standard solution of NaOH. The acid-ba

Question

Procedure for titrating an acid against a standard solution of NaOH. The acid-base indicator, phenolphthalein, is colorless in acidic solution but takes on a pink color in basic solution.

How would the volume of standard solution added change if that solution were Ba(OH)2(aq) instead of NaOH(aq)?

Procedure for titrating an acid against a standard solution of . The acid-base indicator, phenolphthalein, is colorless in acidic solution but takes on a pink color in basic solution.

How would the volume of standard solution added change if that solution were  instead of ?

The volume of standard solution would be half the volume of NaOH(aq) solution.

Sorry for multiple picture! New to this! Thank you for any help!

The volume of standard solution would be twice the volume of NaOH(aq) solution. The volume of standard solution would not change. The volume of standard solution would be 4.3 times greater than the volume of NaOH(aq) solution.

The volume of standard solution would be half the volume of NaOH(aq) solution.

Sorry for multiple picture! New to this! Thank you for any help!

200 ml of acidA few drops of Standard NaOH solution added from burette Solution becomes basic on passing equivalence point, triggering indicator solution added to flask acid-base indicator added Initial volume urette olor change reading color change Final volume reading

Explanation / Answer

Because it is a neutralization reaction, acid and Ba(OH)2 react to form H2O. Basicity of  Ba(OH)2 is twice to that of NaOH, so millimoles produced in the solution by Ba(OH)2 will be twice as much of NaOH. So, the volume of standard solution would be half the volume of NaOH(aq) solution for the pink colouration.

So option D is correct

  

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