IV.2 Titration the of Ca(OH) in 0.010 M Mg(NO)h filtrate Trial 1.5 ml o 35 M a.

Question

IV.2 Titration the of Ca(OH) in 0.010 M Mg(NO)h filtrate Trial 1.5 ml o 35 M a. lici 1.Lo mL .mL (DIV.2c - DIV.2b) (DIV.2a, AlII.2c, and AIV.2a in eq. (13)) (AIV.2b in eq. (11e) (AIV.2c in eq. (14)) N HC c. Solubility (S) 011 M ·00014 M o0014M d. Mean S , 017-7 M Std dey S e. (AIV.2c, d in eq. (16) V.2 Titration the of Ca(OH)2 in 0.010 M Ca(NO)2 filtrate Titration Trial l 2 ml (DV.2c - DV.2b) 038 M o39 M (DV.2a, AIl.2c, and AV.2a in eq. (13)) 019 M . 0192 M c. Solubility (S) (AV.2b in eq. (11c)) olal M Std dev S.o0014LM e. d. Mean S (AV.2c, d in eq. (16) (AV.2c in eq. (14))

Explanation / Answer

The way it is to be done,

Find the decrease in solubility of Ca(OH)2 in presence of Mg(NO3)2.

Calculate the percentage decrease in solubility from pure Ca(OH)2 solubility

This precentage would be used in the second part.

We have solubility data from part II of Ca(OH)2 in Ca(NO3)2 common ion effect.

Obtained above the solubility results.

Now find the percentage similar to the first part, say we have solubility decrease of 5% in first part. So for second part, the correct result would be solubility which is 5% lower than the estimated value.

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