Answer to Question #18532 in Other Chemistry for diya51
NaCl can be prepared from NaOH by reaction with HCl:
NaOH + HCl = NaCl + H2O
4 mL of normal saline content n(NaCl) = V(NaCl) * N(NaCl) = 0.004 L * 1 N = 0.004 mol of NaCl.
This quantity of NaCl can be prepared from equal amountof NaOH.
n(NaOH) = 0.004 mol
m(NaOH) = n(NaOH) * M(NaOH) = 0.004 mol * 40 g/mol = 0.16g
w(NaOH) = m(NaOH) / m(NaOH solution)
m(NaOH solution) = m(NaOH) / w(NaOH) = 0.16 g / 0.01 = 16g
Thus, for preparing 4 mL normal saline solution from 1% NaOH, you should take 16 g of NaOH solution and neutralize it with appropriate amount of HCl (control pH to pH=7). When you finish the neutralization you should evaporate the obtained solution to 4 mL volume.
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