Calculate pH of 0.00424 M aqueous solution of KOH.Solve the given problem.
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Final answer: 11.63
Steps:
1) We have,
Since,KOH is strong base .
![[OH^-] = 0.00424 = 424 * 10^{-5}<br /><br />](https://tex.z-dn.net/?f=+%5BOH%5E-%5D+%3D+0.00424+%3D+424+%2A+10%5E%7B-5%7D%3Cbr+%2F%3E%3Cbr+%2F%3E "[OH^-] = 0.00424 = 424 * 10^{-5}<br /><br />")
2) We know,
pH = 14- pOH
![pH = 14-(-\log([OH^-])) \\ <br /><br />= 14+\log([OH^-] ) <br /><br />\\ <br /><br />= 14+ \log(424*10^{-5}) \\ <br /><br />= 14-5+\log(424) \\ <br /><br />= 9+ 2.63 \\ <br /><br />=11.63](https://tex.z-dn.net/?f=+pH+%3D+14-%28-%5Clog%28%5BOH%5E-%5D%29%29+%5C%5C+%3Cbr+%2F%3E%3Cbr+%2F%3E%3D+14%2B%5Clog%28%5BOH%5E-%5D+%29+%3Cbr+%2F%3E%3Cbr+%2F%3E%5C%5C+%3Cbr+%2F%3E%3Cbr+%2F%3E%3D+14%2B+%5Clog%28424%2A10%5E%7B-5%7D%29+%5C%5C+%3Cbr+%2F%3E%3Cbr+%2F%3E%3D+14-5%2B%5Clog%28424%29+%5C%5C+%3Cbr+%2F%3E%3Cbr+%2F%3E%3D+9%2B+2.63+%5C%5C+%3Cbr+%2F%3E%3Cbr+%2F%3E%3D11.63+ "pH = 14-(-\log([OH^-])) \\ <br /><br />= 14+\log([OH^-] ) <br /><br />\\ <br /><br />= 14+ \log(424*10^{-5}) \\ <br /><br />= 14-5+\log(424) \\ <br /><br />= 9+ 2.63 \\ <br /><br />=11.63")
Hence, pH of 0.00424M solution of KOH is 11.63
Steps:
1) We have,
Since,KOH is strong base .
2) We know,
pH = 14- pOH
Hence, pH of 0.00424M solution of KOH is 11.63
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