<p>1.what would be the pH of a solution prepared by combining equal quantities of NaH2PO4 AND Na2HPO4?</p>
<li>sufficient strong acid is added to a solution containing Na2HPO4 TO NEUTRALIZE ONE-HALF OF IT. WHAT WILL BE THE pH of this solution?explain…</li>
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<p>i kinda figured out second problem but…for first one i have no idea…</p>
<p>any help will be appreciated…</p>
<p>Um. Don't you need concentrations?</p>
<p>Hey, would the pH for the first one even change?
I am not sure.
So for the second one, they didn't give you the Ka?</p>
<p>It(the pH) probably with but you need the concentration</p>
<p>For the first one, you could pretty much ignore the Na2HPO4 because its dissociation is so little, it has nearly no effect on pH. If you know the concentration of the NaH2PO4, you would use that along with its Ka to figure out the pH of the solution.</p>
<p>In fact, HPO4 has such a small Ka that it actually acts like a weak base and takes up some of the H+ ions produced by H2PO4.</p>
<p>Answers to both 1 and 2 are found the same way - you don't need concentration but you do need Ka. In both cases, the final solution is a 50/50 blend of a weak acid and its conjugate base. The pH of any solution of a weak acid and its conjugate is found by pH=pKa+log[base]/[acid]. (the Henderson-Hasselbalch equation) When the solution is 50/50, the equation reduces to pH=pKa because log(1)=0. (This is a classic buffer solution.)</p>