Answer to Question #112269 in Inorganic Chemistry for Delaney Gilder

Question #112269
What volume (in mL) of 0.295 M HCl is required to neutralize 50.0 mL of 0.800 M NaOH?
1
Expert's answer
2020-04-28T12:21:37-0400

Solution:

The neutralization reaction is: NaOH + HCl = H2O + NaCl

According to the chemical equation: n(NaOH) = n(HCl), or:

C(NaOH) × V(NaOH) = C(HCl) × V(HCl)

Substitute the values and solve for V(HCl):

V(HCl) = [C(NaOH) × V(NaOH)] / C(HCl)

V(HCl) = (0.800 M × 50.0 mL) / (0.295 M) = 135.59 mL = 135.6 mL

V(HCl) = 135.6 mL


Answer: 135.6 mL of HCl is required.

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