Answer to Question #71911 in Chemistry for princess

Question

Values for kw at0,50 and 100°c are 1.14×10^-15 , 5.47×10^-14 and 4.9×10^-13 calculate the pH for a neutral solution at each of these temperatures

Answer 1

Answer:
Kw = [H + ][OH - ]
In neutral solution [H] = [OH]
[H + ] = sqrt Kw
pH = - lg [H + ]
1) [H + ] 1 = sqrt (1.14×10 ^-15 ) = 3.4 ·10 ^-8
pH = - lg [3.4 ·10^ -8 ] = 7.4
2) [H + ] 2 = sqrt (5.47×10 ^-14 ) = 2.3 ·10^ -7
pH = - lg [2.3 ·10^ -7 ] = 7.6
3) [H + ] 3 = sqrt (4.9×10^ -13 ) = 7 ·10^ -7
pH = - lg [7 ·10 ^-7 ] = 6.1