We’re being asked to determine the pH after addition of 30.0 mL of 0.300 M HCl(aq)
Let’s first write the balanced reaction between HCl and KOH:
▪ HCl → strong acid
▪ NaOH → (OH- with Group 1A ion) → strong base
▪ the reaction between a strong base and strong acid → no need to create an ICE chart
Balanced reaction: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Let’s first calculate the initial amount of HCl in moles:
moles HCl= 0.009 mol HCl
Calculate the pH of the resulting solution if 30.0 mL of 0.300 M HCl(aq) is added to
40.0 mL of 0.350 M NaOH(aq)
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