I need help with my homework PRE-LAB PROBLEMS:1. Calculate logarithms and antilogarithms. You should be able to do the first three withouta calculator.NumberLog1000log C1o00) = 30.01log C10-2) =-21x 10-4-4log (o-0a)= -1-69log (300 )= 2.47.023003.22. Complete the following table.[H*][OH]pHРОНА,В,N?.1.017

As pH is the measure of acidity and pOH is the measure of basicity. If pH = 7, then solution is neutral, If pH < 7 , then solution is acidic and If pH > 7 , then solution is basic in nature. As we know, pH = -log [H+] ..........(1) pOH = -log [OH -1] ..........(2) Also there is a realtion between pH and pOH pH + pOH = 14 ...........(3) Above three equations are used to complete the table asked in the question. 1.) Now, [H+] = 0.1 (Given in table) pH = -log[H+] = -log (0.1) pH = 1 (acidic) pOH = 14 - pH = 14 - 1 pOH = 13 As, pOH = -log [OH -1] 13 = pOH = -log [OH -1] [OH -1] = 10-132.) [OH -1] = 0.01 ( given in table) pOH = -log [OH -1] = -log [0.01] pOH = 2 pH = 14 - pOH = 14 - 2 pH = 12 ( basic) Also, pH = -log [H+] therefore 12 = pH = -log [H+] [H+] = 10-12 3.) pH = 2 (given in table) Solution is acidic. pH = -log [H+] = 2 , therefore [H+] = 10-2 = 0.01 Also, pOH = 14 - pH = 14 -2 pOH = 12 Now, pOH = -log [OH -1] = 12 [OH -1] = 10-124.) pOH = 7 (given in table) Therefore, pH = 14 - pOH = 14 -7 pH = 7 (neutral) As pH = -log [H+] = 7, therefore [H+] = 10-7 Also, pOH = -log [OH-] = 7 , therefore [OH-] = 10-7 Table is shown in an image below.