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Biology and Chemistry Homework Help
Titration Troubles: Where Did We Go Wrong?
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[QUOTE="Zayn, post: 5999521, member: 640985"] [h2]Homework Statement [/h2] 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? [h2]Homework Equations[/h2] C1V1=C2V2 and unit conversions [h2]The Attempt at a Solution[/h2] [/B] 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! [/QUOTE]
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Titration Troubles: Where Did We Go Wrong?
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