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0.2 mol of HCL will give 0.2 mol of H+ ions (strong acid; so full

ionization in water), same goes for 0.25 mol NaOH (strong alkali) that

will give 0.25 mol. of OH-.

H+ + OH- -> H2O

So effectively there'll be 0.05 mol of OH- after mixing 1 mol of each

HCL & NaOH, which translates to 0.025mol of OH- for every mol of the

mixture.

pH = 14 - log10(OH-) = 12.4

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Q: What is the pH of solution with 0.2 molarity hydrochloric acid with 0.25 molarity sodium hydroxide at the equivalence point?
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