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20.0 mL of 0.10 M hydrochloric acid (HCl) reacts with 20.0 mL of 0.30 M potassium hydroxide (KOH) solution - VCE - SSCE Chemistry - Question 13 - 2005 - Paper 1

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20.0 mL of 0.10 M hydrochloric acid (HCl) reacts with 20.0 mL of 0.30 M potassium hydroxide (KOH) solution. The concentration of potassium ions in the resultant sol... show full transcript

Worked Solution & Example Answer:20.0 mL of 0.10 M hydrochloric acid (HCl) reacts with 20.0 mL of 0.30 M potassium hydroxide (KOH) solution - VCE - SSCE Chemistry - Question 13 - 2005 - Paper 1

Step 1

Calculate moles of HCl

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Answer

To find the moles of HCl, use the formula:

extmoles=extconcentrationimesextvolume ext{moles} = ext{concentration} imes ext{volume}

For HCl: extmolesextHCl=0.10extMimes0.020extL=0.002extmoles ext{moles}_{ ext{HCl}} = 0.10 ext{ M} imes 0.020 ext{ L} = 0.002 ext{ moles}

Step 2

Calculate moles of KOH

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Answer

Similarly, for KOH: extmolesextKOH=0.30extMimes0.020extL=0.006extmoles ext{moles}_{ ext{KOH}} = 0.30 ext{ M} imes 0.020 ext{ L} = 0.006 ext{ moles}

Step 3

Determine the limiting reactant

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Answer

The balanced reaction is:

ightarrow ext{KCl} + ext{H}_2 ext{O} $$ Since 0.002 moles of HCl reacts with 0.002 moles of KOH, HCl is the limiting reactant as less is present.

Step 4

Calculate remaining moles of KOH

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Answer

After reaction: extmolesextKOHremaining=0.0060.002=0.004extmoles ext{moles}_{ ext{KOH remaining}} = 0.006 - 0.002 = 0.004 ext{ moles}

Step 5

Calculate total volume of the solution

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Answer

The total volume after mixing the two solutions is: extTotalVolume=20.0extmL+20.0extmL=40.0extmL=0.040extL ext{Total Volume} = 20.0 ext{ mL} + 20.0 ext{ mL} = 40.0 ext{ mL} = 0.040 ext{ L}

Step 6

Calculate the concentration of potassium ions

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Answer

The concentration of K+ ions is given by: ext{Concentration}_{ ext{K}^+} = rac{ ext{moles}_{ ext{KOH remaining}}}{ ext{Total Volume}} = rac{0.004 ext{ moles}}{0.040 ext{ L}} = 0.10 ext{ M}

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