4 practice questionsNo Description metadataNo Objectives metadataJenny KoenigNo Organisation metadataNo Version metadataNo Date metadataNo Copyright metadataCreated by Course Genie, Single-user Edition.4A_DbothimslrnnoneTrueTrue200True4A_D practice qnsFalseTrueFalseCG2006107185657selftest0regc:/program files/coursegenie/resources/en.lproj/schemes/Standard UTF-84A-C Questions on logarithm notationThe idea of these questions is to get you to become familiar with the notation, so please do not use a calculatorlog_{10}(10000) = 4well donethink about what a would be in 10^{a} = 10000log_{10}(0.01) = -2well donethink about what the number a would be in 10^{a} = 0.01log_{10}10 = 1well donethink about what the number a would be to give 10^{a} = 10log_{10}(1) = 0well donethink about what number a would give 10^{a} = 1log_{10}(10 000 000) = 7well donethink about what number a would give 10^{a} = 10 000 0004D Using logs - the pH scale.Practice calculating pH from [H+] and vice versa.What is the pH of stomach acid which has a hydrogen ion concentration of 0.0125 M?pH = 1.9 (to one decimal place)well doneyou need to calculate log_{10}(0.0125)Milk of magnesia has a hydrogen ion concentration of 3 x 10^{-10} M. pic001.gifpic002.gif Its pH is 9.5 (to one decimal place)well doneyou need to take the log (to the base 10) of 3 x 10^{-10}Fresh milk has a hydrogen ion concentration of 4 x 10^{-7} M. What is its pH (to one decimal place)? 6.4well donetake the log of 4 x 10-7Rainfall in Wisconsin varies from a pH of 4.4 in southeastern Wisconsin to nearly 5.0 in northwestern Wisconsin. This difference in pH of 0.6 units corresponds to what difference in hydrogen ion concentration? pic003.gifpic004.gif
Thanks to the Wisconsin department of natural resources for the data. http://www.dnr.state.wi.us/org/water/fhp/lakes/under/acidity.htmThe hydrogen ion concentration corresponding to pH 4.4 = 40 μM (2 significant figures) the hydrogen ion concentration corresponding to pH 5.0 = 10 μM (2 significant figures). The difference between these concentrations is 30 μM. The concentration of hydrogen ions is 4 times larger at pH 4.4 than pH 5.0.
well doneyou need to antilog -4.4 (to get 40 μM) and -5.0 (to get 10 uM). To get the difference - you subtract (40 - 10 = 30 μM). To find out how much larger you can divide one by the other (40/10 = 4)Anaesthetists have to ensure that a patient is breathing at an appropriate rate since the removal of CO_{2} from the blood is essential in maintaining blood pH within the normal range. In an experiment, an acute change in ventilation caused a drop in blood pH from 7.40 to 7.34. What was the corresponding difference in the concentration of hydrogen ions?
pH 7.4 corresponds to [H^{+}] = 39.8 nM (to one decimal place) pH 7.34 corresponds to [H^{+}] = 45.7 nM (to one decimal place) difference in [H^{+}] = 5.9 nM (to one decimal place)well donecalculate 10^{-7.4} (= 3.98 x 10^{-8} M = 39.8 nM); similarly for 10^{-7.34} (= 45.7 nM). The difference is therefore 45.7 - 39.8 = 5.9 nM)Calculate the pH of a 0.01 M solution of calcium hydroxide. Assume that calcium hydroxide dissociates completely. Remember that [H^{+}] x [OH^{-}] = 10^{-14} M^{2}pH = 12.3 (to one decimal place)well doneremember that calcium is divalent, so calcium hydroxide is Ca(OH)_{2}. Therefore [OH^{-}] = 0.02 M. Therefore [H^{+}] = 10^{-14} / 2 x 10^{-2} = 5 x 10^{-13}. pH = 12.3Calculate the pH of a solution of HCl prepared by diluting 5 ml of 1 M HCl into a total volume of 1 L.pH of resulting solution = 2.3well donethe resulting solution is 0.005 M HCl. The log of that is -2.3. Therefore the pH is 2.3.What volume of 1 M HCl would be required to decrease the pH of one litre of a 10^{-3} M HCl solution by one pH unit? (Ignore the small change in volume)The volume of 1 M HCl required would be 9 ml.well doneIn one litre of 10^{-3} M HCl there are 10^{-3} mol. A solution with a pH 1 unit lower would be 10^{-2} M HCl. This would contain 10^{-2} mol. Therefore we need to add 0.01 - 0.001 mol = 0.009 mol HCl. The volume required would be 0.009 mol / 1 mol.L^{-1} = 0.009 L = 9 ml