Step1 : To convert from pH to ion concentrations,first apply equation 17-1 to calculate [H3O+]. Then make use of water equilibrium to calculate [OH-] Step 2 : We must rearrange equation pH = -log [H+] , in order to solve for concentration Step 3 : log [H+] = -pH and [H+] = 10 -pH Step 4 ; Now substitute and evaluate: [H+] = 10-3.05 = 8.9 x 10-4 M Step 5: To determine the hydroxide ion concentration,we need to rearrange the water equilibrium expression: Kw = 1.0 x 10 -14 = [H3O+]eq [OH-] and [OH-]eq = 1.0 x 10- 14 / [H3O+] now susbtitute and evaluate: = 1.0 x 10- 14 / 8.9 x 10-4 = 1.1 x 10-11 M
Sample protblem 2 : What is the OH- ion concetraion of a solution of NaOH with pH 13.40 ? Solution: Step 1 : First find pOH from the given pH: pH + pOH = 14 pOH = 14 - 13.40 = 0.60 Step 2 : As you have now the value of pOH you can find the OH- ion concentration using formula pOH = -log [OH-] [OH- ] = 10 - pOH = 10 - 0.60 = 0.25 M
No' of molecules divide by avogadro number , 6×6.023×10^23 so (2.2×10^22)÷(6.023×10^23) = 0.03653 moles moles × Molar mass = mass n×Mr=m 0.03653×40 = 1.46 grams
