PLEASEEE ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Before
ID: 965605 • Letter: P
Question
PLEASEEE ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Before doing the problem identify the kind of problem you are dealing with first 1) Determine the pH of 50.00 mL of a 0.015 M solution of HBrO_3. What kind of problem do you think this is? 2) Calculate the pH of a 0.00050 M KOH solution. What kind of problem do you think this is? 3) Calculate the pH of a 0.00050 M Ca(OH)_2 solution. What kind of problem do you think this is? 4) Calculate the pH of a 0.0198 M solution of Phenol acid. What kind of problem do you think this is? 5) Calculate the pH of a 0.085 M solution of aniline, C_6H_5NH_2. What kind of problem do you think this is? 6) Calculate the pH of a 0.0015 M solution of sodium formate, HCOONa. What kind of problem do you think this is? 7) Calculate the pH of a 0.020 M solution of ammonium chloride, NH_4Cl. What kind of problem do you think this is? 8) Calculate the pH of a solution prepared by mixing 100.0 mL of 0.0100 M acetic acid (CH_3COOH) with 175.0 mL of 0.0100 M sodium acetate, CH_3COONa. What kind of problem do you think this is? 9) A solution is prepared by mixing 100.0 mL of 0.100 M acetic acid (CH_3COOH) with 175.0 mL of 0.100 M sodium acetate, CH_3COONa. Determine the pH after the addition of 40.0 mL of 0.01 M HCl. What kind of problem do you think this is? 10) A solution is prepared by mixing 100.0 mL of 0.100 M acetic acid (CH_3COOH) with 175.0 mL of 0.100 M sodium acetate, CH_3COONa. Determine the pH after the addition of 40.0 mL of 0.01 M NaOH. What kind of problem do you think this is? 11) Determine the volume of 0.182 M phosphoric acid, H_3PO_4, needed to titrate 239.86 mL of 0.197 M sodium hydroxide, NaOH, to the equivalence point? What kind of problem do you think this is?Explanation / Answer
Answer: These are problems to calculate the pH and pOH of the given solutions and also of buffer solutions.
answer to 2nd problem: Given 0.00050M KOH, means KOH is a strong electrolyte, hence molarity = concentration
therefore [OH-] = 0.00050 M
to calculate,pH, we use the formula pOH = -log 10[OH-] where pOH gives the basicity of the solution
substituting we get pOH = -log10 [0.00050] ==> pOH = - {log [50 x 10^-5 ]} = - (-log 10^-5 + log 50 ) =
=[ - (-5 + 1.6990)] = (5 -1.6990) = 3.3010 ===> pOH = 3.3010 also we know that pH + pOH = 14
using above equation pH = 14 - pOH ==> pH = 14-3.3010= 10.6990 ==> pH of KOH = 10.6990
ANSWER to 3rd problem: given Ca(OH)2 ---> Ca2+ + 2 OH-
hence according to this problem we are getting 2 (OH-) ions ,therefore its concentration is 2( 0.00050) = 0.001
from above [OH-] = 0.0010 M , using pOH = - log [OH-] , we get pOH = - [log 10 (0.0010)] = - (log 10^-4)
= -(-4) = 4 {since log 10 =1} , hence we get pOH = 4 , again using pH + pOH = 14 , we get pH = 14-4 =10
therefore pH of Ca(OH)2 = 10
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