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Calculate [OH -] and pH for each of the following solutions. (a) 0.0034 M RbOH [OH-] =_________ M pH...

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Chemistry

Calculate [OH -] and pH for each of the followingsolutions.



(a) 0.0034 M RbOH

[OH-] =_________ MpH = _________




(b) 0.0872 g of KOH in 510.0 mL of solution

[OH -] =__________ MpH = __________




(c) 79.8 mL of 0.00719 M Sr(OH)2 diluted to 900mL

[OH -] = __________ MpH = ____________




(d) A solution formed by mixing 64.0 mL of 0.000880 MSr(OH)2 with 36.0 mL of 2.6 x 10-3 MRbOH

[OH -] = ________MpH =__________

Answer & Explanation Solved by verified expert
3.8 Ratings (749 Votes)
a00034 M RbOHRbOH rubidium hydroxide is a strong base and hence it willdissociate completelyso OH 00034 MpOH log OH log00034 247pH 14 pOH 14 247 1153b00872 g of KOH in 5100 mL of solutionMolar mass of KOH 5611 gmolSo 5611 g of KOH 1 mole 00872 g ofKOH    See Answer
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