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ok so 0.3M means 0.3 moles/Liter

so you need to set up a proportion to find the number of moles of NH3 in 50mL not 1000mL

so you do: 0.3mol/1000mL = x mol/50mL

x=0.000015

so the [NH3] is 1.5x10^-5

NH3 is a weak base so it won't dissociate completely so you need its Kb to find the [H+] it will release

pH= -log[H+]

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11y ago
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Q: What is the pH of 50 mL of 0.3 M of ammonia?
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