Reaction:
<span>HCl + NaOH ---> NaCl + H2O
</span><span>1 mole of HCl = 36,5 g
</span><span>1 mole of NaOH = 40g
</span><span>so, according to the reaction:
</span><span>1 mol HCl = 1 mol NaOH
</span>so, we need > 36,5 g HCl (<u>hydrochloric acid</u><span>)
</span><u>
answer: 36,5 g HCl (hydrochloric acid)
</u><span> ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
</span><span>next question.
</span><span>
1 mole of NaCl = 58,5 g
</span><span>1 mole of H2O = 18g
</span>
so, according to the reaction:
1 mole of HCl (36,5 g) <span>----------------- - 1 mole of NaCl (58,5 g)
</span><span>(the same for NaOH)
i
</span>1 mole of HCl<span> (36,5 g) ------------------ 1 mole of H2O (18 g)
</span>(the same for NaOH)
<span>so, this reaction is stechiometric
</span><u>
answer: 58,5 g NaCl i 18g H2O</u>
Answer:
Good question i really dont know sorry
Explanation:
jhjhhshhjjjh
Pretty sure it’s a Lunar Eclipse
Henderson–Hasselbalch equation is given as,
pH = pKa + log [A⁻] / [HA]
-------- (1)
Solution:
Convert Ka into pKa,
pKa = -log Ka
pKa = -log 1.37 × 10⁻⁴
pKa = 3.863
Putting value of pKa and pH in eq.1,
4.29 = 3.863 + log [lactate] / [lactic acid]
Or,
log [lactate] / [lactic acid] = 4.29 - 3.863
log [lactate] / [lactic acid] = 0.427
Taking Anti log,
[lactate] / [lactic acid]
= 2.673
Result:
2.673 M
lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.
