1a.
Calculate the pH of a solution that contains 0.0282 g of NaOH in 1.50 L of water.
Report your answer to TWO places past the decimal.
1b.
Calculate the pH of a solution that contains 0.0261 g of Ba(OH)2 in 2.04 L of water.
Report your answer to TWO places past the decimal.
1a.
molarity = W/G.M.Wt * volume of solution L
= 0.0282/(40*1.5)
= 0.00047M
NaOH(aq) ----------> Na^+ (aq) + OH^-(aq)
0.00047M ------------------------------ 0.00047M
[OH^-] = [NaOH] = 0.00047M
POH = -log[OH^-]
= -log0.00047
= 3.328
PH = 14-POH
= 14-3.328
= 10.67 >>>>answer
molarity = W/G.M.Wt * volume of solution L
= 0.0281/(171*2.04)
= 8.1*10^-5M
Ba(OH)2(aq) ----------> Ba^+ (aq) + 2OH^-(aq)
8.1*10^-5M ------------------------------ 2*8.1*10^-5M
[OH^-] =2 [Ba(OH)2] = 2*8.1*10^-5M = 0.000162M
POH = -log[OH^-]
= -log0.000162
= 3.79
PH = 14-POH
= 14-3.79
= 10.21 >>>>answer
1a. Calculate the pH of a solution that contains 0.0282 g of NaOH in 1.50 L...
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