7.
The negative logarithum of H^+ molar concentration is called PH
PH = -log[H^+]
b. Ba(OH)2 is strong base
Ba(OH)2(aq) ---------------> Ba^2+ (aq) + 2OH^- (aq)
0.03M 2*0.03M
[OH^-] = 2[Ba(OH)2]
[OH^-] = 2*0.03M = 0.06M
POH = -log[OH^-]
= -log0.06
= 1.2218
PH = 14-POH
= 14-1.2218
= 12.7782 >>>>>answer
c.
PKa = -logKa
= -log(4.5*10^-4)
= 3.3467
PH = PKa + log[KNO2]/[HNO2]
= 3.3467 + log0.1/0.085
= 3.3467 + 0.07058
= 3.417
7. (a) Define pH (b) Calculate the pH of 0.030 M Ba(OH)2 solution (c) Calculate the...
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