Calculate the pH of the resulting solution if 19.0 mL of 0.190 M HCl(aq) is added to [Solved]

a. Moles NaOH = 0.0240 L X 0.190 mol/L = 4.56X10^-3 mol NaOH

b) repeat precisely the calculations in (a)

a. Moles NaOH = 0.0240 L X 0.190 mol/L = 4.56X10^-3 mol NaOH

Extra moles NaOH = 4.56X10^-3 3.61X10^-3 = 9.5X10^-4 mol NaOH
Molarity NaOH = 9.5X10^-4 mol / 0.043 L = 0.022 M NaOH
pOH = 1.66
pH = 14.00 pOH = 12.34

b) repeat precisely the calculations in (a)

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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