Question 10 of 13 Calculate the pH when 41.0 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M $CH_3NH_2$ ($K_b = 4.4 \times 10^{-4}$).
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For HBr: moles of HBr = (0.200 mol/L) * (0.041 L) = 0.0082 mol For CH3NH2: moles of CH3NH2 = (0.400 mol/L) * (0.030 L) = 0.012 mol Show more…
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