explain the nature of resulting solution when 20 ml of 0.05 M NaOH is mixed with 10 ml of 0.1 N HCL solution at 25 degree celsius. when 10 ml of this solution is diluted 100 times what is the change in ph

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Arun · Answer

0.05 M NaOH = 0.05 N NaOH as number of equivalents = 1 for NaOH moles of NaOH=0.05 N×20 ml=1 m mole moles of HCl=0.1 N×10 ml=1 m mole Number of moles of acid and base are equal thus solution is neutral.

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explain the nature of resulting solution when 20 ml of 0.05 M NaOH is mixed with 10 ml of 0.1 N HCL solution at 25 degree celsius. when 10 ml of this solution is diluted 100 times what is the change in ph

explain the nature of resulting solution when 20 ml of 0.05 M NaOH is mixed with 10 ml of 0.1 N HCL solution at 25 degree celsius. when 10 ml of this solution is diluted 100 times what is the change in ph

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explain the nature of resulting solution when 20 ml of 0.05 M NaOH is mixed with 10 ml of 0.1 N HCL solution at 25 degree celsius. when 10 ml of this solution is diluted 100 times what is the change in ph

explain the nature of resulting solution when 20 ml of 0.05 M NaOH is mixed with 10 ml of 0.1 N HCL solution at 25 degree celsius. when 10 ml of this solution is diluted 100 times what is the change in ph

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