Chemistry, 24.07.2019 23:30 rubimachuca1020
The weak base ammonia, nh3, and the strong acid hydrochloric acid react to form the salt ammonium chloride, nh4cl. given that the value of kb for ammonia is 1.8×10−5, what is the ph of a 0.289 m solution of ammonium chloride at 25∘c
Answers: 2
![K_{\text{c}} = \dfrac{\text{[NH$_{4}^{+}$][OH$^{-}$]}}{\text{[NH$_3$]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.289-x} = 1.8 \times 10^{-5}\\\\\text{Check for negligibility of }x\\\\\dfrac{0.289 }{1.8 \times 10^{-5}} = 16 000 400\\\\\therefore x \ll 0.289\\\\\begin{array}{rcl}\\\dfrac{x^{2}}{0.289}& = & 1.8 \times 10^{-5}\\\\x^{2}& = & 0.289 \times1.8 \times 10^{-5}\\x^{2}& = & 5.202 \times 10^{-6}\\x & = &2.281 \times 10^{-3}\\\end{array}\\](/tpl/images/0128/9943/ae796.png)
![\text{[H$_{3}$O$^{+}$]}= x \text{ mol$\cdot$L$^{-1}$} = 2.281 \times 10^{-3} \text{ mol$\cdot$L$^{-1}$}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{2.281 \times 10^{-3}} = \boxed{\mathbf{2.64}}\\\text{The pH of the solution is } \boxed{\textbf{2.64}}](/tpl/images/0128/9943/aab51.png)
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