What is the pH of a solution prepared by mixing 50.0 mL of 0.300 M HCl with 450.0 mL of 0.400 M HIO3?

Text Solution

4.749.268.784.63

Answer : C

Solution : `pK_(a)(NH_(4)^(+))=9.26` <br> `:. pK_(b)(NH_(3))=pK_(w)-pK_(a)(NH_(4)^(+))` <br> `=14-9.26=4.74` <br> On adding 15 m mol (50 mL ) `NH_(3)` the resulting solution contains 15 m mol ( in 100 mL ) of `NH_(4)` Cl and 5 m mol ( in 100 mL ) of `NH_(3)`. <br> `:. ["Salt"]=15xx10^(-2)M,["Base"]=5xx10^(-2)M` <br> `:. pOH =pK_(b)+log . (["Salt"])/(["Base"])` <br> `=4.74+log. (15xx10^(-2))/(5xx10^(-2))` <br> `=4.74 +log 3=4.74+0.48=5.22` <br> `pH =14-pOH =14-5.22=8.78`

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• Thread starter Shackleford
• Start date Aug 6, 2008

Answers and Replies

0.714 looks OK

Although it should be more like 0.71, you are given only 2 significant digits of Ka.

&

2

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