# Question #8b799

Apr 24, 2017

pH = 10,5

#### Explanation:

At first yuo must calculate the mole of anphetamine from its MM ( C9H13N=135; C= 12x9 = 108; H= 1x13 = 13; N=14) $m o l = \left(m \frac{g}{M} M\right) = \frac{230}{135} = 1 , 7 {10}^{- 3} m o l$ .$K b = {10}^{-} \left(p K b\right) = {10}^{- 4 , 2} = 6 , 3 {10}^{- 5}$. $\left(O {H}^{-}\right) = \sqrt{C b x K b} = \sqrt{1 , 7 {10}^{- 3} 6 , 3 {10}^{- 5}} = \sqrt{1 , 07 {10}^{- 7}} = 3 , 27 {10}^{-} 4$ $p O H = - L o g \left(O {H}^{-}\right) = - L o g \left(3 , 27 {10}^{-} 4\right) = 3 , 48$ pH = 14 -pOH = 14 -3,48 = 10,52

Apr 24, 2017

At first yuo must calculate the mole of anphetamine from its MM ( C9H13N=135 g/mol; C= 12x9 = 108; H= 1x13 = 13; N=14) $m o l = \left(\frac{m g}{M M}\right) = \frac{230 m g}{135 \frac{g}{m o l}} = 1 , 7 {10}^{- 3} m o l$ .$K b = {10}^{-} \left(p K b\right) = {10}^{- 4 , 2} = 6 , 3 {10}^{- 5}$. $\left(O {H}^{-}\right) = \sqrt{C b K b} = \sqrt{1 , 7 {10}^{- 3} 6 , 3 {10}^{- 5}} = \sqrt{1 , 07 {10}^{- 7}} = 3 , 27 {10}^{-} 4$ $p O H = - L o g \left(O {H}^{-}\right) = - L o g \left(3 , 27 {10}^{-} 4\right) = 3 , 48$
pH = 14 -pOH = 14 -3,48 = 10,52