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Question 24 1 pts Calculate the pH of a 0.417 M solution of Sr(OH)2. Remember that a pH with three decimal places has three s
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Answer #1

Sr(OH)2 (s) ----------> Sr​​​​​2+ (aq) + 2OH- (aq)

Concentration of Sr(OH)2 = 0.417 M

Concentration of OH​​​​​- = 2 × 0.417M

= 0.834M

[OH​​​​​- ] = 0.834 M

pOH = - log (OH​​​​​-​​​​ )

= - log ( 0.834) = 0.078

pH + pOH = 14

pH = 14- 0.078 = 13.92

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