Equal is the answer
Hope this helps
Answer :=11
Thank me later
Answer:
![\boxed{11.13}](https://tex.z-dn.net/?f=%5Cboxed%7B11.13%7D)
Explanation:
The chemical equation is
![\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ +\text{OH}$^{-}$](https://tex.z-dn.net/?f=%5Crm%20NH%24_%7B3%7D%24%20%2B%20%5Ctext%7BH%7D%24_%7B2%7D%24O%20%5C%2C%20%5Crightleftharpoons%20%5C%2C%24%20NH%24_%7B4%7D%5E%7B%2B%7D%24%20%2B%5Ctext%7BOH%7D%24%5E%7B-%7D%24)
For simplicity, let's rewrite this as
![\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$](https://tex.z-dn.net/?f=%5Crm%20B%20%2B%20H%24_%7B2%7D%24O%20%5C%2C%20%5Crightleftharpoons%5C%2C%24%20BH%24%5E%7B%2B%7D%24%20%2B%20OH%24%5E%7B-%7D%24)
1. Initial concentration of NH₃
![\text{[B]} = \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{0.050 mol}}{\text{0.500 L}} = \text{0.100 mol/L}](https://tex.z-dn.net/?f=%5Ctext%7B%5BB%5D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7Bmoles%7D%7D%7B%5Ctext%7Blitres%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.050%20mol%7D%7D%7B%5Ctext%7B0.500%20L%7D%7D%20%3D%20%5Ctext%7B0.100%20mol%2FL%7D)
2. Calculate [OH]⁻
We can use an ICE table to do the calculation.
B + H₂O ⇌ BH⁺ + OH⁻
I/mol·L⁻¹: 0.100 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.100 - x x x
![K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bb%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BBH%7D%5E%7B%2B%7D%5D%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BB%5D%7D%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.100%20-%20x%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D)
Check for negligibility:
\![\dfrac{0.100 }{1.8 \times 10^{-5}} = 5600 > 400\\\\ x \ll 0.100](https://tex.z-dn.net/?f=%5Cdfrac%7B0.100%20%7D%7B1.8%20%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%205600%20%3E%20400%5C%5C%5C%5C%20x%20%5Cll%200.100)
3. Solve for x
![\dfrac{x^{2}}{0.100} = 1.8 \times 10^{-5}\\\\x^{2} = 0.100 \times 1.8 \times 10^{-5}\\\\x^{2} = 1.80 \times 10^{-6}\\\\x = \sqrt{1.80 \times 10^{-6}}\\\\x = \text{[OH]}^{-} = 1.34 \times 10^{-3} \text{ mol/L}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.100%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.100%20%5Ctimes%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%201.80%20%5Ctimes%2010%5E%7B-6%7D%5C%5C%5C%5Cx%20%3D%20%5Csqrt%7B1.80%20%5Ctimes%2010%5E%7B-6%7D%7D%5C%5C%5C%5Cx%20%3D%20%5Ctext%7B%5BOH%5D%7D%5E%7B-%7D%20%3D%201.34%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctext%7B%20mol%2FL%7D)
4. Calculate the pH
![\text{pOH} = -\log \text{[OH}^{-}] = -\log(1.34 \times 10^{-3}) = 2.87\\\\\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.87 = \mathbf{11.13}\\\\\text{The pH of the solution at equilibrium is } \boxed{\mathbf{11.13}}](https://tex.z-dn.net/?f=%5Ctext%7BpOH%7D%20%3D%20-%5Clog%20%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20-%5Clog%281.34%20%5Ctimes%2010%5E%7B-3%7D%29%20%3D%202.87%5C%5C%5C%5C%5Ctext%7BpH%7D%20%3D%2014.00%20-%20%5Ctext%7BpOH%7D%20%3D%2014.00%20-%202.87%20%3D%20%5Cmathbf%7B11.13%7D%5C%5C%5C%5C%5Ctext%7BThe%20pH%20of%20the%20solution%20at%20equilibrium%20is%20%7D%20%5Cboxed%7B%5Cmathbf%7B11.13%7D%7D)