Question

Calculate the molar concentrations of H+ and OH in solutions that have the following pOH values. Enter values in scientific notation, for example 0.0010 would be 1.0E-3. x Incorrect. Did you remember to include the negative signs in the pH and pOH equations? a) 12.07 [OH): 71.4E-13 6.9E-2 [H+]: Correct. b) 7.00 [OH-]: 1.0E-7 [H+]: 1.0E-7 3.1.0-7 SHOW SOLUTION x Incorrect. Did you remember to include the negative signs in the pH and pOH equations? c) 10.89 (1.9E-13 (OH): 5.2E-2 [H+]: X Incorrect. Did you remember to include the negative signs in the pH and pOH equations? d) 4.95 [OH): 76.7E-12 1.5E-3 [H+]: X Incorrect. Did you remember to include the negative signs in the pH and pOH equations? e) 3.82 [OH": 14.5E-11 Twth. 2.2E-4 2.2E-4 ihti

I can't figure these questions out. Help would be appreciated!

Answer 1

I © @ civen, pot = 112.07 We know that pH + pot = 14:00 pH = 1.93] pH = 1.93 - log (+] = 193 > [t] = 12E-2 pOH = 12.07 -log [OF] = 1207 CH] = 8.5E-13 | pot = 7 - PH= 14:00-7:00 (pH = 7:00] pH=7.00 -log[+] = 700 [(+1] = 100E-7 | рбH = 1:17 -log[h] = 700 / COH] = 1.0 E-7 ] pOH = 10189 - pH=14.00-10.89 >> [pH = 3:11] pH = 3.11 -LogCH] = 3 11 (+] = 7.8E-4 po= 10.89 -log Cont) = 10189 - (07)= 1-3E-11 pot: 4.95 pH= 14:00-4.95 (pH = 9:05] pH = 9:05 -log CH J = 9:05 [H] = [8-9E-10 pot=4.95 -log[ot] = 4.95 (OH] = /11E-5 pOH = 3.82 pH = 14:00-3:82 pH = 10.18) pH = 10:18 -log (ett ] = 10:18 (+] = 6:6E-11 ] pot = 3.82 -log [OH) = 3:82 [OF] = 1.5E-4 |