");
cont.document.writeln("");
cont.document.close();
cont.focus();
}
function startit(){
clear()
totl++
document.forms[0].total.value=totl
getspecies()
}
function getspecies(){
first=Math.floor(19*Math.random())
if(Math.random()>.5){
option=1
cmpd[0]=compound[first][0]
cmpd[2]=compound[first][1]
k1=compound[first][2]
acidk=0
}
else{
option=2
cmpd[1]=compound[first][1]
cmpd[3]=compound[first][0]
k2=1e-14/compound[first][2]
basek=1
}
second=first
while(second==first){second=Math.floor(37*Math.random())
if(option==1){
if(second<19){
cmpd[1]=compound[second][1]
cmpd[3]=compound[second][0]
k2=1e-14/compound[second][2]
basek=2
}
else{
cmpd[1]=compound[second][0]
cmpd[3]=compound[second][1]
k2=compound[second][2]
basek=0
}
}
else
if(second<19){
cmpd[0]=compound[second][0]
cmpd[2]=compound[second][1]
k1=compound[second][2]
acidk=0
}
else{
cmpd[0]=compound[second][1]
cmpd[2]=compound[second][0]
k1=1e-14/compound[second][2]
acidk=1
}
}
thereaction=cmpd[0]+" + "+cmpd[1]+" => " +cmpd[2]+" + "+cmpd[3]
K=k1*k2/1e-14
showK[0]=number(k1,2,2)
showK[1]=number(k2,2,2)
showK[2]=number(K,2,2)
buildquestion(option,first,second)
}
function buildquestion(){
thequest="Calculate the value of the equilibrium constant for the following proton transfer reaction: "+thereaction
showquest(thequest)
buildanswer()
}
function buildanswer(){
theanswer="Krxn=Ka*Kb/Kw In this case, "
if(option==1){
theanswer+=("Ka for "+cmpd[0]+"="+showK[0])
theanswer+=" "
basek==0?theanswer+=("Kb for "+cmpd[1]+"="+showK[1]):theanswer+=(cmpd[1]+" is the conjugate base of " + compound[second][0]+". Its Kb=Kw/Ka=1*10-14/"+number(compound[second][2],2,2)+"="+showK[1])
}
else{
acidk==0?theanswer+=("Ka for "+cmpd[0]+"="+showK[0]):theanswer+=(cmpd[0]+" is the conjugate acid of " + compound[second][0]+". Its Ka=Kw/Kb=1*10-14/"+number(compound[second][2],2,2)+"="+showK[0]);
theanswer+=" "+(cmpd[1]+" is the conjugate base of " + compound[first][0]+". Its Kb=Kw/Ka=1*10-14/"+number(compound[first][2],2,2)+"="+showK[1])
}
theanswer+=" "
theanswer+=(showK[0]+"*"+showK[1]+"/10-14="+showK[2])
}
function showquest(thequest){
if(ie4){
qspot.innerHTML=thequest
}
else{
document.qspot.document.open()
document.qspot.document.write(thequest," ")
document.qspot.document.close()
}
}
function showanswer(){
if(attempts>=3){
if(ie4){
ansspot.innerHTML=theanswer
}
else{
document.ansspot.document.open()
document.ansspot.document.write(theanswer)
document.ansspot.document.close()
}
}
}
function clearansspot(){
if(ie4){
ansspot.innerHTML=""
}
else{
document.ansspot.document.open()
document.ansspot.document.clear()
document.ansspot.document.close()
}
}
function answer(ans){
if(tried==1)alert("You've done this one. Get a new problem.");
else{
attempts++
if(tried==2){totl++
document.forms[0].total.value=totl
}
if(ans>.95*K & ans<1.05*K ){
document.forms[0].results.value="correct"
corrt++
document.forms[0].correct.value=corrt
tried=1
}
else {document.forms[0].results.value="incorrect"
tried=2}
}}
Ks for acid-base reactions
This page randomly generates problems involving proton transfer equilibria where all of the species are weak. In order to use the page, you need a
table of acid/base Ks. To start,press "New Problem" and a question will appear to the right of the button. Determine the value of the
equilibrium constant for the reaction shown, enter it in the cell and press "Check Answer". Results appear in the table.
If you use scientific notation enter the number is the form 2.3e5.
If you get a problem "incorrect", you should redo it and recheck your
answer.
If you miss a problem three times, pressing the "Show Answer" button will cause a complete solution to appear.