Answers: 2
![\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}}](/tpl/images/0312/0435/12713.png)
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![\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613](/tpl/images/0312/0435/4476a.png)
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.![\displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}](/tpl/images/0312/0435/b3b8e.png)
.![\displaystyle [\text{BH}^{+}] = \frac{n}{V} = \frac{0.380}{0.5} = 0.760\;\text{mol}\cdot\text{dm}^{-3}](/tpl/images/0312/0435/ab546.png)
.![\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991](/tpl/images/0312/0435/c762d.png)
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If a buffer contains 1.05m b and 0.750m bh+ has the ph of 9.5. what would be the ph after 0.005mol o...
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