What is the pH of a solution in which 43 mL of 0.16 M NaOH is...

What is the pH of a solution in which 43 mL of 0.16 M NaOH is added to 16 mL of 0.140 HCl M ?

pH =

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Expert Solution

HCl + NaOH ----> NaCl + H2O

moles of HCl = M*Volume

= 0.140*0.016

= 0.00224 moles

moles of NaOH = M*Volume

= 0.16 *0.043

= 0.00688 moles

Moles of NaOH is more than that of HCl.

moles of NaOH left = 0.00688 - 0.00224

= 0.00464

pOH = -log(0.00464)

= 2.33

pH = 14 - 2.33

= 11.67

queen_honey_blossom answered 1 year ago

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