Answer to Question #113586 in General Chemistry for Emma

Question

calculate the final pH when 120.0mL of 0.50M HI is added to 850 mL of 0.05M HClO4

Answer 1

Solution.

$V(HI) = 120.0mL=0.120L;$

$M(HI) =0.50M;$

$V(HClO_4) =850mL=0.850L;$

$M(HClO_4)=0.05M;$

$\nu(HI)=V(HI)\sdot M(HI); \nu(HI)=0.120L\sdot0.50M=0.06mol;$

$\nu(HClO_4)=V(HClO_4)\sdot M(HClO_4);$

$\nu(HClO_4)=0.850L\sdot0.05M=0.0425mol;$

$\nu_{final}=0.06mol+0.0425mol=0.1025mol;$

$[H_3O^+]=\dfrac{\nu_{final}}{V_{final}}; [H_3O^+]=\dfrac{0.1025mol}{0.970L}=0.1057M;$

$pH=-lg[H_3O^+]=-lg(0.1057)=lg9.46=0.976$ ;

Answer: $pH=0.976.$