Sulfamic acid (HSO3NH2) is a strong monoprotic acid which can be used to standardize a strong base...

Question

agent_a · Answer

$$HSO_3NH_2+KOH  \rightarrow H_2O+K ^{+}+SO_3NH_2^{+}$$

A 0.177-g sample of $$HSO_3NH_2$$required 19.4mL of an aqueous solution of KOH. 
What is the Molarity of the KOH solution?

agent_a · Answer

I don't think my solution is right:

$$0.177gHSO_3NH_2\times \frac{ 1 mol HSO_3NH_2 }{ 97.1g/mol }=0.001822863028molHSO_3NH_2$$
$$\frac{ 0.001822863028molHSO_3NH_2 }{ 19.4mLtimes \frac{ 1L }{ 1000mL } }=0.093962011743MHSO_3NH_2$$

and then I realized that I need the molarity of KOH, not $$HSO_3NH_2$$ ....

So I started with the mass of KOH which is 56.108g KOH, and then I got stuck.

aaronq · Answer

when you have a monoprotic acid and a base that only abstracts 1 proton (or creates 1 OH), you can use:  $M_{acid}V_{acid}=M_{base}V_{base}$

for your example you'd use  $n_{acid}=M_{base}V_{base}$ instead.


or you can work through it the long way (stoichiometrically) to arrive at the same answer.

agent_a · Answer

But I still need to calculate the Molarity, yes?

Because from what I understand, I only know that $$V_2=0.0194L$$and I'd have to solve for the Molarity of the acid and Base as well. 

I used: $$0.177gHSO_3NH_2 \times \frac{ 1molHSO_3NH_2 }{ 97.1g/molHSO_3NH_2 }=0.001822975673molHSO_3NH_2$$

but then I did not proceed because I don't know if I should use 19.4mL of KOH to calculate the molarity of $$HSO_3NH_2$$ ....

aaronq · Answer

you don't need the molarity of the acid, since the moles are equal to the M*V product, you can use what i posted earlier.