Calculate the pH of the resulting solution if 25.0 mL" style="position: relative;" role="presentation">25.0 mL of 0.250 M HCl(aq)" style="position: relative;" role="presentation">0.250 M HCl(aq) is included to 35.0 mL" style="position: relative;" role="presentation">35.0 mL of 0.250 M NaOH(aq)." style="position: relative;" role="presentation">0.250 M NaOH(aq).

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pH=" style="position: relative;" role="presentation">pH=

Calculate the pH of the resulting solution if 25.0 mL" style="position: relative;" role="presentation">25.0 mL

of 0.250 M HCl(aq)" style="position: relative;" role="presentation">0.250 M HCl(aq) is added to 15.0 mL" style="position: relative;" role="presentation">15.0 mL of 0.350 M NaOH(aq)." style="position: relative;" role="presentation">0.350 M NaOH(aq).

pH=" style="position: relative;" role="presentation">pH=

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1)

Given:

M(HCl) = 0.25 M

V(HCl) = 25 mL

M(NaOH) = 0.25 M

V(NaOH) = 35 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.25 M * 25 mL = 6.25 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.25 M * 35 mL = 8.75 mmol

We have:

mol(HCl) = 6.25 mmol

mol(NaOH) = 8.75 mmol

6.25 mmol of both will certainly react

remaining mol of NaOH = 2.5 mmol

Total volume = 60.0 mL

= mol of base continuing to be / volume

= 2.5 mmol/60.0 mL

= 4.167*10^-2 M

use:

pOH = -log

= -log (4.167*10^-2)

= 1.3802

use:

PH = 14 - pOH

= 14 - 1.3802

= 12.6198

2)

Given:

M(HCl) = 0.25 M

V(HCl) = 25 mL

M(NaOH) = 0.35 M

V(NaOH) = 15 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.25 M * 25 mL = 6.25 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.35 M * 15 mL = 5.25 mmol

We have:

mol(HCl) = 6.25 mmol

mol(NaOH) = 5.25 mmol

5.25 mmol of both will certainly react

staying mol of HCl = 1 mmol

Total volume = 40.0 mL

= mol of acid remaining / volume

= 1 mmol/40.0 mL

= 2.5*10^-2 M

use:

pH = -log

= -log (2.5*10^-2)

= 1.6021

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