Question

what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m na2so4? what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m sr(oh)2?

Answer 1

The concentration of hydroxide ions when NaOH is added to Na$_{2}$SO$_{4}$ is 0.074 M, The concentration of hydroxide ions when NaOH is added to Sr(OH)$_{2}$ is 0.356 M

When NaOH is added to Na$_{2}$SO$_{4}$

OH from NaOH  is X

0.250 M × 50.0 ml = X × 170 ml

X = 0.074 M of OH

Since there are no OH ions in Na$_{2}$SO$_{4}$ so the ions in NaOH is only considered

When NaOH is added to Sr(OH)$_{2}$

OH from NaOH  is X$_{1}$

0.250 M × 50.0 ml = X$_{1}$ × 170 ml

X$_{1}$ = 0.074 M

OH ion from Sr(OH)$_{2}$ is X    $_{2}$

0.200 M × 120 ml = X  $_{2}$ × 170 ml

X$_{2}$ = 0.141 M for one Sr(OH)

For Sr(OH)$_{2}$ = 2 × X$_{2}$ = 0.282 M of OH

Now we need to add both OH ions

Total concentration of hydrogen ions = 0.074+0.282

= 0.356 M

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