How to prepare 100 mL of lead nitrate Pb(NO3)2 ?
1 answer:
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Answer:
0,07448M of phosphate buffer
Explanation:
sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:
Na₂HP + HCl ⇄ NaH₂P + NaCl <em>(1)</em>
NaH₂P + HCl ⇄ H₃P + NaCl <em>(2)</em>
The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:
0,01862L of HCl×= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.
The concentration is:
= <em>0,07448M of phosphate buffer</em>
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I hope it helps!