What is the [OH-] of solution with pH of 3.4? (A) 4 x 0.0001M
(B) 4 x 1/100000000000
(C) 2.5 x 0.0001M
(D) 2.5 x 1/100000000000

(A) 4×0.0001M

Explanation:

[OH-] = 10^-3.4 which is equivalent to 4×0.0001M

2.5 x 10^ -11 (D)

Explanation:

[H+] = 10^ -3.4 = 3.98 x 10^ -4

[OH-] = (1 x 10^ -14) ÷ [H+] = 2.5 x 10^ -11

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol