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a 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution after the addition of 50.0 ml of lioh.

User Iodbh
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The pH of the solution after addition of 50.0ml is 0.1003

pH is known as the negative logarithm of the hydrogen ion concentration in a solution.

In the given condition we have 100 ml of sample of 0.18 molar of HclO4 titrated with 0.27m LioH.

pH = log[H+]

= log [Hclo4]

Mole of Hclo4 0.1 L * 0.18 m = 0.018 M

mol LiOH = 0.1 L * 0.27 m = 0.027 M

total volume = 0.1L + 0.1L = 0.2L

HClO4] = (0.018-0.0081) /0.2 L

= -1.26 billion

PH = -㏒ -1.26

= 0.1003

So, the pH of the solution after adding 100 ml of liquid is 0.1003

User Bioneuralnet
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