Calculate the ph of the resulting solution if 27.0 ml of 0.270 m hcl (aq is added to 37.0 ml of 0.270 m naoh (aq

The reaction between NaOH and HCl is as follows

NaOH + HCl - - - > NaCl + H₂O

for neutralisation, H⁺ ions react with an equivalent amount of OH⁻ ions.

Number of NaOH moles reacted = 0.270 M/1000 mL/L x 37 mL = 0.00999 mol

number of HCl moles reacted = 0.270 M/1000 mL x 27 mL = 0.00729 mol

HCl reacts with NaOH in 1:1 molar ratio

Number of excess NaOH moles remaining - 0.00999 - 0.00729 = 0.0027 mol

total volume of solution = 37 mL + 27 mL = 64 mL = 0.064 L

Since there's excess OH⁻ ions, we can calculate pOH value first

pOH = - log [OH⁻]

[OH⁻] = 0.0027 mol / 0.064 L = 0.042 mol/L

pOH = - log (0.042 M)

pOH = 1.37

by knowing pOH we can calculate pH using the following equation;

pH + pOH = 14

pH = 14 - 1.37

pH = 12.63