suppose you had an experimental error of 10% in your measurement of calcium ioda

Question

suppose you had an experimental error of 10% in your measurement of calcium iodates molar solubility. would the percent error in your Ksp be less than, more than or equal to 10% difference, explain why. suppose you had an experimental error of 10% in your measurement of calcium iodates molar solubility. would the percent error in your Ksp be less than, more than or equal to 10% difference, explain why. suppose you had an experimental error of 10% in your measurement of calcium iodates molar solubility. would the percent error in your Ksp be less than, more than or equal to 10% difference, explain why.

Explanation / Answer

Percentage uncertainity is also called relative uncertainity.

Given the relative error in the measurement of solubility (S) of calcium iodate [Ca(IO3)2] is 10%.

Hence solubility = (S+10%) M

------------------- Ca(IO3)2(s) -------- > Ca2+(aq) + 2IO3-(aq)

Eqm.con(M): ---------------------------- (S+10%), 2(S+10%)

Hence Ksp = [Ca2+(aq)]x[IO3-(aq)]2 = (S+10%) x [2(S+10%)]2 = (S+10%) x [2S+10%]2

We get the Ksp value by multiplying the solubilities.

While multiplying two quantities, their relative uncertainities are added up in the final multiplication value.

Hence while calculating the value of Ksp the relative uncertainities will be added up to give 10% + 10% + 10%

= 30 %

Hence the Ksp will have 30% error

Hence the percentage error value is more than 10% (answer)

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