Find pH of the solution that results from the addition of 26.00 ml of 0.200M NaOH to 50.0 ml of 0.100M HCL?

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Answer 1 · 2018-03-19T18:06:53

We interrogate the chemical reaction....

#NaOH(aq) +HCl(aq) rarr NaCl(aq) + H_2O(l)#

And thus there is 1:1 equivalence between the acid and base...

#"Moles of NaOH"=0.200*mol*L^-1xx26.00xx10^-3*L-=5.20xx10^-3*mol.#

#"Moles of HCl"=0.100*mol*L^-1xx50.00xx10^-3*L-=5.00xx10^-3*mol.#

And there are thus #2.0xx10^-4*mol# #NaOH(aq)# EXCESS in a solution whose volume is the sum of the individual volumes, i.e. #76.0*mL#...

And so #[NaOH]=(2.0xx10^-4*mol)/(76xx10^-3*mol)=2.63xx10^-3*mol*L^-1#...

#pOH=-log_10(2.63xx10^-3)=2.58#...

And since #pOH+pH=14# under standard conditions....

#pH=14-2.58=11.4#

The elevated #pH# is consistent with the stoichiometric excess of hydroxide ions.....