Next step, multiply both sides by (.01-x)
.012(.01-x)=.01x+x^2
gather terms, put in standard binomial form, use the quadratic equation, and solve for x.
Your question actually bothers me. You are asking an algebra 1 question in order to solve a very complicated chemistry question. Are you in over your head in chem math? You may want to discuss this with an academic counselor.
k2=[H][SO4]/[HSO4]
.012=(.01+x)(x)/(.01-x)
Solve for x, and calculate pH where H is .01+x
When I am solving for x does it become
.012= .01x + x^2/ .01 -x.. what is the next step of solving for x?
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