Answer:
Explanation:
Using H-H equation for the NaH2O4/Na2HPO4 buffer:
pH = pKa + log [Na2HPO4] / [NaH2PO4]
<em>Where pKa of this buffer is 7.198, [Na2HPO4] = 0.058M = moles [NaH2PO4] = 0.042M = moles</em>
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Replacing:
pH = 7.198 + log [0.058M] / [0.042M]
<em>Initial pH = 7.34</em>
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Now, NaOH reacts with NaH2PO4 as follows:
NaOH + NaH2PO4 → Na2HPO4 + H2O
That means, the moles of NaOH are additional moles of Na2HPO4 and subtracted moles of NaH2PO4.
The moles of NaOH added:
1.0mm³ = 1.0x10⁻³L * (10mol / L) = 0.01 moles
<em>NaH2PO4:</em>
0.042mol - 0.010mol = 0.032moles
<em>Na2HPO4:</em>
0.058mol + 0.010mol = 0.068moles
Replacing in H-H equation:
pH = 7.198 + log [0.068M] / [0.032M]
<em>Final pH = 7.53 </em>
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Change in pH = 7.53 - 7.34
The change in pH is 0.19 units
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