100.0 mL of 3.00 M HCl (in water) is added to 50.0 mL of 5.00 M...

Question

Question

100.0 mL of 3.00 M HCl (in water) is added to 50.0 mL of 5.00 M NaOH in a beaker. What is

present in the beaker?

science chemistry

Answer 1

Answer #1

HCl + NaOH --------------> NaCl + H2O

millimoles of NaOH = 50.0 x 5.00 = 250

millimoles of HCladded = 100.0 x 3.0 = 300

300 - 25 = 50.0 millimoles NaOH left in the solution

total volume = 100 + 50 = 150 mL

[NaOH] = 50 / 150 = 0.33 M

resulting NaOH = 0.33 M

Answer 2

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