150 mL of 0.20 M H2SO4 is mixed with 100 mL of 0.30 M NaOH and 100 mL of 0.10 M Ca(OH)2 (all strong, volumes additive, CaSO4 solubility ignored). Calculate the pH of the final solution (two decimal places).

JEE Advanced levelNumerical answerDifficulty 3 of 3eq148

150 mL of 0.20 M $H X_{2} SO X_{4}$ is mixed with 100 mL of 0.30 M NaOH and 100 mL of 0.10 M $Ca (OH) X_{2}$ (all strong, volumes additive, $CaSO X_{4}$ solubility ignored). Calculate the pH of the final solution (two decimal places).

Numerical answer type, tolerance 0.006.

Acid = 150

\times

0.20

\times

2 = 60.0 meq. Base = 100

\times

0.30 = 30.0 meq (NaOH) plus 100

\times

0.10

\times

2 = 20.0 meq (

Ca (OH) X_{2}

) = 50.0 meq. Excess

H X^{+}

= 10.0 meq in 350 mL, so

[H X^{+}]

= 0.02857 M and pH = 1.54.