You create solutions of HI and NaOH with concentrations of 1.24 and 0.9,respecti

Question

You create solutions of HI and NaOH with concentrations of 1.24 and 0.9,respectively. If you titrate 10.0 mL of the HI solution with the NaOH base you have created, at what volume do you expect to see the equivalence point? mL NaOH 13.8 If the actual mL used is 15.1, what was the actual concentration of the base assuming the acid concentration was correct? Actual [NaOH]0.82 If the actual mL used is 15.1, what was the actual concentration of the acid assuming the base concentration was correct? Actual [HI] 1.36 What is the error or in each case? %Err (Actual . Given)/Actual x 100% %error [NaOH] = -27.5 %error [HI] 44.7 Submit Answer Save Progress Practice Another Version

Explanation / Answer

The first 3 answers are correct according to the image so you want to calculate the percent error, the formula has been given already

Lets solve this for NaOH,

Actual concentration of NaOH = 0.82

Expected concentration of NaOH = 0.9

% error = (Actual - Expected) / Actual * 100

% error = (0.82 - 0.9) / 0.82 =- 0.09756 * 100 = - 9.75 %

% error [HI]

Actual Hi = 1.36

expected = 1.24

% error = (1.36 - 1.24) / 1.36 =0.088

0.088 * 100 = 8.82 %

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