Note @gung's question; it matters. I will assume that the treatment was the same for every tank in the treatment group.
If you can argue the variance would be equal for the two groups (which you would typically assume for a two sample t-test anyway), you can do a test. You just can't check that assumption, no matter how badly violated it might be.
If $\bar{x}$ is the mean of the treatment and $\bar{y}$ is the mean of the control, and both are from normal distributions with variance $\sigma^2$, then $\bar{x}-\bar{y}$ will have mean $\mu_x - \mu_y$ and variance $\sigma^2 (1/n_x + 1/n_y)$ irrespective of whether one of the $n$'s is 1.
So when $n_y$ is 1,
$$ \frac{(\bar{x}-\bar{y})}{s_x\sqrt{1/n_x+1}} $$
(where $s_x$ is the standard deviation computed from the treatments) will be $t$-distributed with $n_x - 1$ degrees of freedom under the null.
You may notice that with the best available estimate of $\sigma$, $s_x$ used for $s_p$, this is exactly like the ordinary two-sample t-test formula with $n_y$ set to 1.
With sample sizes so small, this will be somewhat sensitive to distributional assumptions, however.
If you're prepared to make different assumptions, or want to test equality of some other population quantity, that may be possible.
So all is not lost... but for next time, where possible, it's generally better to have at least some replication in both groups.