What is #pH# of a solution that is prepared from #3.50xx10^-3mol# #NaOH(aq)#, and #5.95xx10^-4mol# #HCl(aq)# in a #1*L# volume of solution? Is the resultant solution alkaline or acidic?

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Answer 1 · 2017-11-15T19:25:41

#pH=11.46#

We gots MOLAR quantities of #3.50xx10^-3*mol# #NaOH# and #5.95xx10^-4*mol# #HCl#....

...and clearly, these react 1:1 to leave EXCESS #NaOH#....

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

And thus #[NaOH]=(3.50xx10^-3*mol-5.95xx10^-4*mol)/(1.00*L)#

#=(2.91xx10^-3)/(1.0*L)=2.91xx10^-3*mol*L^-1#

Now in an aqueous solution, #14=pH+pOH#..

....and #pOH=-log_10(2.91xx10^-3)=-(-2.54)=2.54#

#pH=14-pOH=14-2.54=11.46#, which value betokens an ALKALINE solution.....

What is #pH# of a solution that is prepared from #3.50xx10^-3*mol# #NaOH(aq)#, and #5.95xx10^-4*mol# #HCl(aq)# in a #1*L# volume of solution? Is the resultant solution alkaline or acidic?