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k2=[H][SO4]/[HSO4] .012=(.01+x)(x)/(.01-x) Solve for x,
Calculate the pH of the following aqueous solution:
1.00 mol/L sulfuric acid, H2SO4(aq) H2SO4 ==> H^+ + HSO4^- 100% HSO4^- ==>
2 answers
asked by
Sarah
1,407 views
Calculate the pH of the following aqueous solutions:
1.00 mol/L sulfuric acid H+ = 1.00 M for the first ionization THe second one
1 answer
asked by
Ross
535 views
DrBob has given you the equation to use:
k2 = [H^+][SO4^-]/[HSO4^-] and the values to substitute [H^+] = 0.01 + x [SO4^-2] = x
3 answers
asked by
lyne
543 views
Calculate the molar solubility of BaSO4 in a solution in which [H3O+] is 2.5 M.
I can almost solve this on my own. If you know
1 answer
asked by
Jay
1,160 views
1- Calculate the equilibrium concentration for H+, HSO4- and SO4^2- , given a .150M H2SO4 solution. the Ka for HSO4- is
1 answer
asked by
JOJO
802 views
Complete these Brønsted-Lowry reactions:
HSO4- + H+ <--> HSO4- +OH- <-->
1 answer
asked by
Melissa
1,183 views
When the following redox equation is balanced with smallest whole number coefficients, the coefficient for the hydrogen sulfate
1 answer
asked by
Admir
2,424 views
Identify the two conjugate acid-base pairs in each of the following:
A) H2PO4^-(aq) + HSO4^-(aq) = H3PO4(aq) + SO4^2-(aq) B)
1 answer
asked by
Robyn
1,256 views
H2SO4(aq)+H2O(l)→H3O+(aq)+HSO4−(aq)
solve
1 answer
asked by
jessica
754 views
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
1 answer
asked by
lyne
494 views