The #Ka# of phosphoric acid, #H_3PO_4#, is #7.6 x 10^-3# at 25#"^o#C. For the reaction #H_3PO_4(aq) -> H_2PO_4(aq) + H^+(aq)# #DeltaH# = 14.2 kJ/mol. What is the Ka of #H_3PO_4# at 60#"^o#C?

1 Answer
Feb 12, 2017

Answer:

#"Ka" = 4.2 * 10^-3#

Explanation:

Use the van't Hoff equation:

#ln("K2"/"K1") = ∆"Hº"/R (1/"T1" - 1/"T2") #

#ln ("K2"/"7.6*10^-3") = "-14,200 J"/8.314 (1/298 - 1/333) #

#ln ("K2"/"7.6*10^-3" ) = -1708 (0.00035) #

#ln(" K2"/"0.0076") = -0.598#

#"Apply log rule" a = logb^(b^a)#

-0.598 = #ln (e^(-0.598)) = ln(1/e^0.598)#

Multiply both sides with e^0.598

#K2e^0.598# = 0.0076

#\frac{Ke^{0.598}}{e^{0.598}}=\frac{0.0076}{e^{0.598}}#

#K2 = \frac{0.0076}{e^{0.598}} = 4.2*10^-3#

#"K2" = 4.2* 10^-3#