1. Identify oxidation states and electron change

We need the reaction in basic medium:

• $CrO_4^{2-} \to Cr(OH)_4^-$
• $S_2O_3^{2-} \to SO_4^{2-}$

2. Reduction half-reaction for chromium

Oxidation state of Cr in $CrO_4^{2-}$ is $+6$.

In $Cr(OH)_4^-$: Let oxidation state of Cr be $x$. $x+4(-1)=-1 \Rightarrow x=+3$

So chromium is reduced from $+6$ to $+3$, i.e. it gains 3 electrons.

Balancing in basic medium: $CrO_4^{2-}+4H_2O+3e^- \rightarrow Cr(OH)_4^-+4OH^-$

3. Oxidation half-reaction for thiosulfate

In $S_2O_3^{2-}$, the two sulfur atoms are not equivalent, but their oxidation states are commonly taken as:

• one sulfur: $+5$
• other sulfur: $-1$

In $SO_4^{2-}$, sulfur is $+6$.

So total increase in oxidation number for the two sulfur atoms: $(-1 \to +6) = +7$ $(+5 \to +6) = +1$ Total increase $= 8$

Hence, one mole of $S_2O_3^{2-}$ loses 8 electrons.

Balanced oxidation half-reaction in basic medium: $S_2O_3^{2-}+10OH^- \rightarrow 2SO_4^{2-}+5H_2O+8e^-$

4. Equalize electrons

• Chromium half-reaction involves $3e^-$
• Thiosulfate half-reaction involves $8e^-$

LCM of 3 and 8 is 24.

Multiply:

• chromium half by 8
• thiosulfate half by 3

So stoichiometric ratio is: $8\,CrO_4^{2-} : 3\,S_2O_3^{2-}$

Thus, $\frac{n(CrO_4^{2-})}{n(S_2O_3^{2-})}=\frac{8}{3}$

5. Calculate moles of thiosulfate

Volume of $S_2O_3^{2-}$ solution $=40\text{ mL}=0.040\text{ L}$

Molarity $=0.25\text{ M}$

$n(S_2O_3^{2-})=0.25\times 0.040=0.010\text{ mol}$

6. Calculate required moles of chromate

$n(CrO_4^{2-})=\frac{8}{3}\times 0.010=0.026666\ldots\text{ mol}$

7. Calculate required volume of $0.154$ M chromate solution

$V=\frac{n}{M}=\frac{0.026666\ldots}{0.154}=0.17316\text{ L}$

Convert to mL: $0.17316\text{ L}=173.16\text{ mL}$

Rounded to nearest integer: $\boxed{173}$

8. Comparison with stored answer

Stored correct answer = $173$

Our derived answer matches the stored answer.