- #1

Zayn

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## Homework Statement

We had a titration lab where we took 1.0 g of NaOH and added it to 50 mL of water. We then took that solution and added 10 mL of it to 500 mL of water, producing a 510 mL solution in total (this became the titrant). We then took 0.4 g of potassium hydrogen phthalate and added it to 50 mL of water (this was the analyte).

Our group filled the burette 4 times and didn't get any colour change in the phenolphthalein. Where did we go wrong?

## Homework Equations

C1V1=C2V2 and unit conversions

## The Attempt at a Solution

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I calculated that you need 199.7 mL of NaOH to titrate the KHP; can someone please tell me whether I'm right or not?

This is how I calculated it

(0.4 g KHP) * (1 mol KHP / 204.22 g KHP) = 0.019587 mol KHP and since it's monoprotic also 0.0019587 mol NaOH

next the NaOH

1 g NaOH * (1 mol NaOH / 39.997 g NaOH) = 0.025002 mol NaOH

Now using C1V1=C2V2,

(0025002 mol NaOH)/(50 mL*1L/1000mL) <--- this is the initial concentration, C1

10 mL * 1L / 1000 mL <----- this is the initial volume, V1

the final volume is 510 mL NaOH or 0.510 L

rearranging and solving, I got 0.009805 M for the concentration

C = n/V

V = n/C

V = (0.0019587 mol NaOH)/(0.009805 mol/L NaOH) = 0.1997 L or 199.7 L

Any help would be appreciated. Thanks!

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